Lectures on the Algebraic Theory of Fields - Tata Institute of ...

<strong>Lectures</strong> on theAlgebraic Theory of FieldsByK.G. RamanathanTata Institute of Fundamental Research, Bombay1954

IntroductionThere are notes of course of lectures on Field theory aimed at providingthe beginner with an introduction to algebraic extensions, algebraicfunction fields, formally real fields and valuated fields. These lectureswere preceded by an elementary course on group theory, vectorspaces and ideal theory of rings—especially of Noetherian rings. Aknowledge of these is presupposed in these notes. In addition, we assumea familiarity with the elementary topology of topological groupsand of the real and complex number fields.Most of the material of these notes is to be found in the notes ofArtin and the books of Artin, Bourbaki, Pickert and Van-der-Waerden.My thanks are due to Mr. S. Raghavan for his help in the writing ofthese notes.K.G. Ramanathan

Contents1 General extension fields 11 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . 12 Adjunctions . . . . . . . . . . . . . . . . . . . . . . . . 33 Algebraic extensions . . . . . . . . . . . . . . . . . . . 54 Algebraic Closure . . . . . . . . . . . . . . . . . . . . . 95 Transcendental extensions . . . . . . . . . . . . . . . . 122 Algebraic extension fields 171 Conjugate elements . . . . . . . . . . . . . . . . . . . . 172 Normal extensions . . . . . . . . . . . . . . . . . . . . 183 Isomorphisms of fields . . . . . . . . . . . . . . . . . . 214 Separability . . . . . . . . . . . . . . . . . . . . . . . . 245 Perfect fields . . . . . . . . . . . . . . . . . . . . . . . 316 Simple extensions . . . . . . . . . . . . . . . . . . . . . 357 Galois extensions . . . . . . . . . . . . . . . . . . . . . 388 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . 463 Algebraic function fields 491 F.K. Schmidt’s theorem . . . . . . . . . . . . . . . . . . 492 Derivations . . . . . . . . . . . . . . . . . . . . . . . . 543 Rational function fields . . . . . . . . . . . . . . . . . . 674 Norm and Trace 751 Norm and trace . . . . . . . . . . . . . . . . . . . . . . 752 Discriminant . . . . . . . . . . . . . . . . . . . . . . . 82v

8 1. General extension fields9every irreducible polynomial f (x) in k[x] there exists an extension fieldin which f (x) has a root.Let now g(x) be any polynomial in k[x] and f (x) an irreducible factorof g(x) in k[x]. Let K be an extension of k in which f (x) has a rootξ. Let in Kg(x)=(x−ξ) λ ψ(x).Thenψ(x)∈k[x]. We again take an irreducible factor ofψ(x) andconstruct K ′ in whichψ(x) has a root. After finite number of steps wearrive at a field L which is an extension of k and in which g(x) splitscompletely into linear factors. Letα 1 ,...,α n be the distinct roots ofg(x) in L. We call k(α 1 ,...,α n ) the splitting field of g(x) in L.Obviously (k(α 1 ,...,α n ) : k)≤n!We have therefore the importantTheorem 3. If k is a field and f (x)∈k[x] then f (x) has a splitting fieldK and (K : k)≤n!, n being degree of f (x).It must be noted however that a polynomial might have several splittingfields. For instance if D is the quaternion algebra over the rationalnumber fieldΓ, generated by 1, i, j, k thenΓ(i),Γ( j),Γ( f ),Γ(k) are allsplitting fields of x 2 + 1 inΓ[x]. These splitting fields are all distinct.Suppose k and k ′ are two fields which are isomorphic by means ofan isomorphismσ. Thenσcan be extended into an isomorphism ¯σ ofk[x] on k ′ [x] by the following prescription∑ ∑¯σ( a i x i )= (σa i )x i a i ∈ k σa i ∈ k ′Let now f (x) be a polynomial in k[x] which is irreducible. Denoteby f ¯σ (x), its image in k ′ [x] by means of the isomorphism ¯σ. Then f ¯σ (x)is again irreducible in k ′ [x]; for if not one can by means of ¯σ −1 obtain anontrivial factorization of f (x) in k[x].Let nowαbe a root of f (x) over k andβaroot of f ¯σ (x) over k ′ .Thenk(α)≃k[x]/( f (x)), k ′ (β)≃k ′ [x]/( f ¯σ (x))Letτbe the natural homomorphism of k ′ [x] on k ′ [x]/( f σ (x)). Considerthe mappingτ·σ on k[x]. Sinceσis an isomorphism, it follows

4. Algebraic Closure 9thatτ·σ is a homomorphism of k[x] on k ′ [x]/(F σ (x)). The kernel of thehomomorphism is the set ofϕ(x) in k[x] such that( )ϕ σ (x)∈ f σ (x).This set is precisely ( f (x)). Thus10k[x]/( f (x))≃k ′ [x]/( f ¯σ (x))By our identification, the above fields contain k and k ′ respectivelyas subfields so that there is an isomorphismµof k(α) on k ′ (β) and therestriction ofµto k isσ.In particular if k=k ′ , then k(α) and k(β) are k− isomorphic i.e., theyare isomorphic by means of an isomorphism which is identity on k. Wehave thereforeTheorem 4. If f (x)∈k[x] is irreducible andαandβare two roots ofit (either in the same extension field of k or in different extension fields),k(α) and k(β) are k− isomorphic.Note that the above theorem is false if f (x) is not irreducible in k[x].4 Algebraic ClosureWe have proved that every polynomial over k has a splitting field. For agiven polynomial this field might very well coincide with k itself. Supposek has the property that every polynomial in k has a root in k. Thenit follows that the only irreducible polynomials over k are linear polynomials.We make now theDefinition. A fieldΩis algebraically closed if the only irreducible polynomialsinΩ[x] are linear polynomials.We had already defined the algebraic closure of a field k containedin a field K. Let us now make theDefinition. A fieldΩ/k is said to be an algebraic closure of k if

10 1. General extension fields1) Ω is algebraically closed2) Ω/k is algebraic. 11We now prove the importantTheorem 5. Every field k admits, upto k-isomorphism, one and onlyone algebraic closure.Proof. 1) Existence. Let M be the family of algebraic extensions K αof k. Partially order M by inclusion. Let{K α } be a totally orderedsubfamily of M. Put K= ⋃ K α for K α in this totally ordered family.αNow K is a field; forβ 1 ∈ K,β 2 ∈ K meansβ 1 ∈ K α for someαandβ 2 ∈ K β for someβ. Thereforeβ 1 ,β 2 in K α or K β whichever is largerso thatβ 1 +β 2 ∈ K. Similarlyβ 1 β −12∈ K. Now K/k is algebraicsince everyλ ∈ K is in some K α and so algebraic over k. ThusK∈M and so we can apply Zorn’s lemma. This proves that M has amaximal elementΩ.Ω is algebraically closed; for if not let f (x) be anirreducible polynomial inΩ[x] andρaroot of f (x) in an extensionΩ(ρ) ofΩ. Then sinceΩ/k is algebraic. Ω(ρ) is an element of M.This contradicts maximality ofΩ. ThusΩis an algebraic closure ofk.2) Uniqueness. Let k and k ′ be two fields which are isomorphic bymeans of an isomorphismσ. Consider the family M of triplets{(K, K ′ ,σ) α } with the property 1) K α is an algebraic extension ofk, K α ′ of k ′ , 2)σ α is an isomorphism of K α on K α ′ extendingσ. Bytheorem 4, M is not empty. We partially order M in the followingmanner□12(K, K ′ ,σ) α < (K, K ′ ,σ) βIf 1) K α ⊂ K β , K α ′ ⊂ K′ β , 2)σ β is an extension ofσ α . Let{K, K ′ ,σ) α }be a simply ordered subfamily. Put K= ⋃ K α , K ′ = ⋃ K α ′ ,α αThese are then algebraic over k and k ′ respectively.Define ¯σ on K by

4. Algebraic Closure 11¯σx=σ α xwhere x∈K α . (Note that every x∈K is in some K α in the simplyordered subfamily). It is easy to see that ¯σ is well - defined. Supposex∈K β and K β ⊂ K α thenσ α is an extension ofσ ρ and soσ α x=σ β x.This proves that ¯σ is an isomorphism of K on K ′ and extendsσ. Thusthe triplet (K, K ′ , ¯σ) is in M and is an upper bound of the subfamily.By Zorn’s lemma there exists a maximal triplet (Ω,Ω ′ ,τ). We assertthatΩis algebraically closed; for if not letρbe a root of an irreduciblepolynomial f (x)∈Ω[x]. Then f ζ (x)∈Ω ′ [x] is also irreducible. Letρ ′be a root of f τ (x). Thenτcan be extended to an isomorphism ¯τ ofΩ(ρ)onΩ ′ (ρ). Now (Ω(ρ), ¯τ is in M and hence leads to a contradiction. ThusΩ is an algebraic closure of k,Ω ′ of k ′ andτan isomorphism ofΩonΩ ′ extendingσ.In particular if k=k ′ andσthe identity isomorphism, thenΩandΩ ′ are two algebraic closures of k andτis then a k-isomorphism.Out theorem is completely demonstrated.Let f (X) be a polynomial in k[x] and K= k(α 1 ,...,α n , a splittingfield of f (x), so thatα 1 ,...,α n are the distinct roots of f (x) in K. Let K ′be any other splitting field andβ 1 ,...β m the distinct roots of f (x) in K ′ .LetΩbe an algebraic closure of K andΩ ′ of K ′ . ThenΩandΩ ′ are twoalgebraic closures of k. There exists therefore an isomorphismσofΩonΩ ′ which is identity on k. LetσK= K 1 . Then K 1 = k(σ α1 ,...,σ αn ).Sinceα 1 ,...α n are distinctσα 1 ...,σα n are distinct and are roots off (x). Thus K 1 is a splitting field of f (x) inΩ ′ . This proves that 13K ′ = K 1 .β 1 ,...,β m are distinct and are roots of f (x) inΩ ′ . We have m=nandβ i =σα 2 in some order. Therefore the restriction ofσto K is anisomorphism of K on K ′ . We haveTheorem 6. Any two splitting fields K, K ′ of a polynomial f (x) in k[x]are k− isomorphic.Let K be a finite field of q elements. Then q=P n where n is aninteger≥ 1 and p is the characteristic of K. Also n=(K :Γ),Γbeing

12 1. General extension fieldsthe prime field. Let K ∗ denote the abelian group of non-zero elementsof K. Then K ∗ being a finite group of order q−1,α q−1 = 1for allα∈K ∗ . Thus K is the splitting field of the polynomialinΓ[x]. It therefore followsx q − xTheorem 7. Any two finite fields with the same number of elements areisomorphic.A finite field cannot be algebraically closed; for if K is a finite fieldof q elements and a∈ K ∗ the polynomial∏f (x)= x (X− b)+ab∈K ∗is in K[x] and has no root in K.5 Transcendental extensions14We had already defined a transcendental extension as one which containsat least on transcendental element.Let K/k be a transcendental extension and Z 1 ,...,Z n any n elementsof K. Consider the ring R=k[x 1 ,... x n ] of polynomials over k in nvariables. Let Y be the subset of R consisting of those polynomialsf (x 1 ,... x n ) for whichf (Z 1 ,...Z n )=0.Y is obviously an ideal of R. If Y = (o) we say that Z 1 ,...Z n arealgebraically independent over k. If Y (o), they are said to bealgebraically dependent. Any element of K which is algebraic overk(Z 1 ,...,Z n ) is therefore algebraically dependent on Z 1 ,...Z n .We now define a subset S of K to be algebraically independent overk if every finite subset of S is algebraically independent over k. If K/kis transcendental there is at least one such non empty set S .

5. Transcendental extensions 13Let K/k be a transcendental extension and let S , S ′ be two subset ofK with the propertiesi) S algebraically independent over kii) S ′ algebraically independent over k(s)Then S and S ′ are disjoint subsets of K and S US ′ are algebraicallyindependent over k. That S and S ′ are disjoint is trivially seen. Let nowZ 1 ,...Z m ∈ S and Z ′ 1 ,...Z′ n ∈ S′ be algebraically dependent. This willmean that there is a polynomial f ,f=f (x 1 ,..., x m+n )in m+n variables with coefficients in k, such thatf (Z 1 ,...,Z m , Z ′ 1 ,...Z′ n )=0.Now f can be regarded as a polynomial in x m+1 ,..., x m+n with coefficientsin k(x 1 ,..., x m ). If all these coefficients are zero then Z 1 ,...Z m ,Z1 ′,...Z′ n are algebraically independent over k. If some coefficient is 0, then f (Z 1 ,...Z m , x m+1 ,..., x m+n ) is a non zero polynomial overk(S ) which vanishes for x m+1 = Z1 ′,... x m+n= Z n ′ which contradicts thefact that S ′ is algebraically independent over k(S ). Thus f = o identicallyand our contention is proved. 15The converse of the above statement is easily proved.An extension field K/k is said to be generated by a subset M of K ifK/k(M) is algebraic. Obviously K itself is a set of generators. A subsetB of K is said to be a transcendence base of K if1) B is a set of generators of K/k2) B algebraically independent over k.If K/k is transcendental, then, it contains algebraically independentelements. We shall prove that K has a transcendence base. Actuallymuch more can be proved as in

14 1. General extension fieldsTheorem 8. Let K/k be a transcendental extension generated by S andA a set of algebraically independent elements contained in S . Thenthere is a transcendence base B of K withA⊂ B⊂SProof. Since S is a set of generators of K, K/k(S ) is algebraic. Let Mbe the family of subsets A α of K with1) A⊂A α ⊂ S2) A α algebraically independent over k.□16The set M is not empty since A is in M. Partially order M by inclusion.Let{A α } be a totally ordered subfamily. Put B 0 = ⋃ αA α . ThenB o ⊂ S . Any finite subset of B o will be in some A α for largeαand soB o satisfies 2) also. Thus using Zorn’s lemma there exists a maximalelement B in M. Every element x of S depends algebraically on B forotherwise BU x will be in M and will be larger than B. Thus k(S )/k(B)is algebraic. Since K/k(S ) is algebraic, it follows that B satisfies theconditions of the theorem.The importance of the theorem is two fold; firstly that every set ofelements algebraically independent can be completed into a transcendencebase of K and further more every set of generators contains abase.We make the following simple observation. Let K/k be an extension,Z 1 ,...Z m , m elements of K which have the property that K/k(Z 1 ,...Z m )is algebraic, i.e., that Z 1 ,...,Z m is a set of generators. If Z∈Kthen Zdepends algebraically on Z 1 ,...,Z m i.e., k(Z, Z 1 ,...,Z m )/k(Z 1 ,...,Z m )is algebraic. We may also remark that if in the algebraic relation connectingZ, Z 1 ,...Z m , Z 1 occurs then we can say thatk(Z, Z 1 ,...Z m )/k(Z, Z 2 ,...,Z m )is algebraic which means that Z, Z 2 ,...Z m is again a set of generators.We now prove the

5. Transcendental extensions 15Theorem 9. If K/k has a transcendence base consisting of a finite numbern of elements, every transcendence base has n elements.Proof. Let Z 1 ,...,Z n and Z1 ′,...,Z′ m be two transcendence bases consistingof n and m elements respectively. If nm let n

16 1. General extension fields18A transcendental extension K/k is said to be purely transcendentalif there exists a base B with K= k(B). Note that this does not mean thatevery base has this property. For instance if k(x) is the field of rationalfunctions of x then x 2 is also transcendental over x but k(x 2 ) is a propersubfield of k(x).Let K= k(x 1 ,... x n ) and K ′ = k(x ′ 1 ,... x′ n ) be two purely transcendentalextensions of dimension n. Consider the homomorphismσ definedbyσ f (x 1 ,... x n )= f (x ′ 1 ,..., x′ n)Where f (x 1 ,..., x n )∈k[x 1 ,..., x n ]. It is then easy to see that this isan isomorphism of K on K ′ . This provesTheorem 11. Two purely transcendental extensions of the same dimensionn over k k-isomorphic.This theorem is true even if the dimension is infinite.

Chapter 2Algebraic extension fields1 Conjugate elementsLetΩbe an algebraic closure of k and K an intermediary field. LetΩ ′ 19be an algebraic closure of K and so of k. Then there is an isomorphismτ ofΩ ′ onΩwhich is trivial on k. The restriction of this isomorphism toK gives a fieldτK inΩwhich is k-isomorphic to K. Conversely supposeK and K ′ are two subfields ofΩwhich are k-isomorphic. SinceΩis acommon algebraic closure of K and K ′ , there exists an automorphismofΩwhich extends the k-isomorphism of K and K ′ . Thus1) Two subfields K, K ′ ofΩ/k are k-isomorphic if and only if there existsa k-automorphismσ ofΩsuch thatσK=K ′ .We call two such fields K and K ′ conjugate fields over k.We define two elementsω,ω ′ ofΩ/k, to be conjugate over k if thereexists a k-automorphismσ ofΩsuch thatσω=ω ′The automorphisms ofΩwhich are trivial on k form a group and sothe above relation of conjugacy is an equivalence relation. We cantherefore put elements ofΩinto classes of conjugate elements overk. We then have17

18 2. Algebraic extension fields2) Each class of conjugate elements over k contains only a finite numberof elements.20Proof. Let C be a class of conjugate elements andω∈C. Let f (x)be the minimum polynomial ofωin k[x]. Letσbe an automorphismofΩ/k. Thenσω∈C. Butσω is a root of f σ (x)= f (x). Also ifω ′ ∈ C thenω ′ =σω for some automorphismσofΩ/k. In that caseσω=ω ′ is again a root of f (x). Thus the elements in C are all roots ofthe irreducible polynomial f (x). Our contention follows. □Notice that ifα,βare any two roots, lying inΩ, of the irreduciblepolynomial f (x), then k(α) and k(β) are k-isomorphic. This isomorphismcan be extended into an automorphism ofΩ. ThusTheorem 1. To each class of conjugate elements ofΩthere is associatedan irreducible polynomial in k[x] whose distinct roots are all theelements of this class.Ifα∈Ω we shall denote by C α the class ofα. C α is a finite set.2 Normal extensionsSuppose K is a subfield ofΩ/k andσan automorphism ofΩ/k. LetσK⊂K. We assert thatσK=K. For letα∈ K and denote by ¯C α thesetC α ∩ KSinceσK⊂ K we haveσα∈ K soσα∈ ¯C α . Thusσ ¯C α ⊂ α21¯C α is a finite set andσis an isomorphism of K into itself.Thusσ ¯C α = ¯C αwhich meansα∈σK. Thus K=σK.We shall now study a class of fields K⊂Ω/k which have the propertyσK⊂K,

2. Normal extensions 19for all automorphismsσ ofΩ/k. We shall call such fields, normal extensionsof k inΩ.Let K/k be a normal extension of k andΩalgebraic closure of kcontaining K. Letα∈Kand C α the class ofα. We assert that C α ⊂ K.For ifβis an element of C α , there is an automorphismσ ofΩ/k forwhichβ=σα. SinceσK⊂K, it follows thatβ∈K. Now any elementαin K is a root of an irreducible polynomial in k[x]. Since all the elementsof C α are roots of this polynomial, it follows that if f (x) is an irreduciblepolynomial with one root in K, then all roots of f (x) lie in K.Conversely let K be a subfield ofΩ/k with this property. Letσbean automorphism ofΩ/k andα/K. Letσbe an automorphism ofΩ/kandα∈K. Let C α be the class ofα. Since C α ⊂ K,σα∈K. Butαisarbitrary in K. ThereforeσK⊂ Kand K is normal. Thus theTheorem 2. Let k⊂K⊂Ω. ThenσK= K for all automorphismsσofΩ/k⇐⇒ every irreducible polynomial f (x)∈k[x] which has one rootin K has all roots in K.Let f (x) be a polynomial in k[x] and K its splitting field. LetΩbean algebraic closure of K. Letα 1 ,...,α n be the distinct roots of f (x) in 22Ω. ThenK= k(α 1 ,...,α n )Letσbe an automorphism ofΩ/k.σα j =α j for some j. Thusσtakes the setα 1 ,...,α n onto itself. Since every element of K is a rationalfunction ofα 1 ,...,α n , it follows thatσK⊂ K. Thusi) The splitting field of a polynomial in k[x] is a normal extension ofk.Let{K α } be a family of normal subfields ofΩ/k. Then ⋂ K α isαtrivially normal. Consider k( ⋃ K α ). This again is normal since forαany automorphismσ ofΩ/k.⎛⋃ ⋃ ⋃σk⎜⎝K α⎞⎟⎠⎛⎜⎝ ⊂ k σK α⎞⎟⎠⎛⎜⎝ ⊂ k K α⎞⎟⎠ααα

20 2. Algebraic extension fieldsLet now{ f α (x)} be a set of polynomials in k[x] and K α their splittingfields, then K( ⋃ K α ) is normal. Also it is easy to see thatα⎛⋃L=k⎜⎝K α⎞⎟⎠is the intersection of all subfields ofΩ/k in which every one of thepolynomials f α (x) splits completely. Thusii) if{ f α (x)} is a set of polynomials in k[x], the subfield ofΩgeneratedby all the roots of{ f α (x)} is normal.We also haveα23iii) If K/k is normal and k⊂L⊂Kthen K/L is also normal.For ifσis an L- automorphism ofΩ, thenσis also a k-automorphismofΩand soσK⊂ K.The k−automorphisms ofΩform a group G(Ω/k). From what wehave seen above, it follows that a subfield K ofΩ/k is normal if andonly ifσK=Kfor everyσ∈G(Ω/k). Now a k− automorphismof K can be be extended into an automorphism ofΩ/k, becauseevery such automorphism is an isomorphism of K inΩ. It thereforefollowsiv) K/k is normal if and only if every isomorphism of K inΩ/k is anautomorphism of K over k.As an example, letΓbe the field of rational numbers and f (x)=x 3 − 2. Then f (x) is irreducible inΓ[x]. Letα= 3√ 2 be one of its roots.Γ(α) is of degree 3 overΓand is not normal since it does not containρ whereρ= −1+√ −32. However the fieldΓ(α,ρ) of degree 6 overΓisnormal and is the splitting field of x 3 − 2.If K is the field of complex numbers, consider K(z) the field of rationalfunction of 2 over K. Consider the polynomial x 3 −z in K(z)[x]. Thisis irreducible. Letω=z 3 1 be a root of this polynomial. Then K(z)(ω) isof degree 3 K(z) and is the splitting field of the polynomial x 3 − z.

3. Isomorphisms of fields 213 Isomorphisms of fieldsLet K/k be an algebraic extension of k and W any extension of K and soof k. A mappingσof K into W is said to be k− linear if forα,β∈K 24σ(α+β)=σα+σβσα∈Wand ifλ∈k,σ(λα)=λσα. Ifσis a k−linear map of K intoW we defineασ forαin W by(ασ)β=ασβforβ∈ K. This again is a k−linear map and so the k−linear maps of Kinto W form a vector space V over W.A k−isomorphismσ of K into W is obviously a k−linear map andsoσ∈V. We shall say, two isomorphismsσ,τ of K into W (trivial onk) are distinct if there exists at least oneω∈K,ω0 such thatσωτωLet S be the set of mutually distinct isomorphisms of K into W. Wethen haveTheorem 3. S is a set of linearly independent elements of V over W.Proof. We have naturally to show that every finite subset of S is linearlyindependent over W. Let on the contraryσ 1 ,...,σ n be a finite subset ofS satisfying a non trivial linear relation∑α i σ i = 0iα i ∈ W. We may clearly assume that no proper subset ofσ 1 ,...,σ nis linearly dependent. Then in the above expression allα i are differentfrom zero. Letωbe any element of K. Then∑α i σ i ω=0i□

22 2. Algebraic extension fieldsIf we replaceωbyωω ′ we get, sinceσ ′ i s are isomorphisms, 25∑α i σ i ω ′ .σ i ω=oifor everyω∈ K. This means thatσ i ,...,σ n satisfy another linear relation∑α i σ i ω ′ .σ i ω=oiSince the isomorphisms are mutually distinct, we can chooseω ′ inK in such a way thatσ 1 ω ′ σ n ω ′We then get from the two linear relations, the expression∑n−1( αiαn −σ iω ′ α)iσ n ω ′ σ i = o.α ii=1This relation is non trivial since the coefficient ofσ 1 is differentfrom zero. This leads to a contradiction and our theorem is proved.Suppose dim V

3. Isomorphisms of fields 23mapping. It is uniquely determined by its effects onω 1 ,...,ω n . Putα i =σω i and letτbe given by∑τ=σ− α i σ iThenτ(ω j )=σω j∑iα i σ i (ω j )=oso thatτ=o. Our contention isestablished.From this we obtain the very importantCorollary . If (K : k)

24 2. Algebraic extension fields28Let nowσbe any automorphism ofΩ/k. LetσK=K (i) . Then ittakesα (1) into a rootα (i j) of f σ (x)= f σ i(x) whereσ i is the isomorphismwhich coincides withσon K. Thus since every isomorphismof K(α) over k comes from an automorphism ofΩ/k, our contention isestablished.Let now K= k(ω 1 ,...,ω n ) and K i = k(ω 1 ,...,ω i ) so that K o = kand K n = K. Let K i have over K i−1 exactly P i distinct K i−1 -isomorphisms.Then K/k has exactly p 1··· p n distinct k-isomorphisms inΩ.Hence2) If K⊃ L⊃k be a tower of finite extensions and K has ever L,n distinct L-isomorphisms inΩand L has over k, m distinct k-isomorphismsthen K has over k precisely mn distinct k-isomorphisms.In particular let (K : k)

4. Separability 25For, the minimum polynomial ofωover L divides that over k.2) ω∈Ω separable over k⇔k(ω)/k has inΩ(k(ω) : k) distinct k- 29isomorphisms.Let now K= k(ω 1 ,...,ω n ) and letω 1 ,...ω n be all separable over k.Put K i = k(ω 1 ,...,ω) so that K o = k and K n = K. Now K i−1 (ω i ) andω i is separable over K i−1 so that K i over K i−1 has exactly (K i : K i−1 )distinct K i−1 - isomorphisms. This proves that K has over k(K n : K n−1 )...(K 1 : K o )=(K n : K o )=(K : k)distinct k-isomorphisms. If thereforeω∈K, Then by previous articlek(ω) has exactly (k(ω) : k) distinct isomorphisms and henceωis separable over k. Conversely if K/k is finite and every elementof K is separable over k, then K/k has exactly (K : k) distinct k-isomorphisms. Hence3) (K : k) < ∞, K/k has (K : k) distinct k-isomorphisms⇔everyelement of K is separable over k.Let us now make theDefinition. A subfield K ofΩ/k is said to be separable over k if everyelement of K is separable over k.From 3) and the definition, it follows that4) K/k is separable⇔ for every subfield L of K with (L : K)

26 2. Algebraic extension fields6) If{K α } is a family of separable subfields ofΩthen2) ⋂ K α and b)k( ⋃ K α ) are separable.α αa) follows easily because every element of K α is separable over k.b) follows since every element of k( ⋃ K α ) is a rational function of aαfinite number of elements and as each of these is separable the resultfollows from 3).7) Let K/k be any extension-not necessarily algebraic. The set L ofelements of k separably algebraic over K is a field.This is evident. We call L the separable closure of k in K.We had already defined an algebraic elementωto be inseparable ifits minimum polynomial has repeated roots. Let us study the natureof irreducible polynomials.Let f (x)=a o + a 1 x+···+a n x n be an irreducible polynomial in k[x].If it has a rootω∈Ω which is repeated, thenωis a root off 1 (x)=a 1 + 2a 2 x+···+na n x n−1 .Thus f (x)| f 1 (x) which can happen only ifia i = o, i=1,...,n.31Let k have characteristic zero. Then ia i = o⇒a i = 0 that is f (x) isa constant polynomial. Thus8) Over a field of characteristic zero, every non constant irreduciblepolynomial has all roots distinct.Let now k have characteristic po. if pχi then ia i = o⇒a i = o.Thus for f 1 (x) to be identically zero we must have a i = o for pχi. Inthis casef (x)=a o + a p x p +···or that f (x)∈k[x p ]. Let e be the largest integer such that f (x)∈k[x p ] but not in k[x pe+1 ]. Consider the polynomialφ(y) withφ(x pe )=

4. Separability 27f (x). Thenφ(y) is irreducible in k[y] andφ(y) has no repeated roots.Letβ 1 ,...,β t be the roots ofφ(y) inΩ. Thenf (x)=(x pe −β 1 )···(x pe −β t ).Thus n=t· p e . The polynomial x Pe −β i has inΩall roots identicalto one of them sayα i . Thenx Pe −β i = x Pe −∆α pei= (x−α i ) peThusf (x)={(x−α 1 )···(x−α t )} peMoreover sinceβ 1 ...β t are distinct, α 1 ,...,α t are also distinct.Hence9) Over a field of characteristic po, the roots of an irreducible polynomialare repeated equally often, the multiplicity of a root being p e ,e≥o.It is important to note that (x−α 1 )...(x−α t ) is not a polynomial in 32k[x] and t is not necessarily prime to p.We call t the reduced degree of f (x) (or of any of its roots ) and p e ,its degree of inseparability. ThusDegree of: Reduced degree X-degree of inseparabilityIfω∈Ω then we had seen earlier that k(ω)/k has as many distinctisomorphisms inΩas there are distinct roots inΩof the minimumpolynomial ofωover k. If we call the reduced degree of k(ω) as thekreduced degree ofω we have10) Reduced degree ofω = Number of distinct roots of the minimumpolynomial ofωover k.We may now call a polynomial separable if and only if every root ofit inΩis separable. In particular if f (x)∈k[x] is irreducible thenf (x) is separable if one root of it is separable.

28 2. Algebraic extension fieldsLetω∈Ω and f (x) the minimum polynomial ofωin k[x]. If t=reduced degree ofω, thenf (x)={(x−ω 1 )...(x−ω t )} pen=t− p e . Letω 1 =ω. Considerω pe1 =β 1. Thenf (x)=(x pe −β 1 )···(x pe −β t )andβ 1 ,...β t are separable over k. Consider the field k(β 1 ) which isa subfield of k(ω).β being of degree t over k, (k(β 1 ) : k)=t. Thismeans that(k(ω) : k(β))= p e .33But the interesting fact to note is that k(ω) has over k(β) only theidentity isomorphism or that k(ω) is fixed by every k(β)-automorphismofΩ/k(β).Also since every element of k(ω) is a rational function ofωover k(β),it follows thatλ pe ∈ k(β)for everyλ∈k(ω). Thus the integer e has the property that for everyλ∈k(ω),λ pe ∈ k(β) and there is at least oneλ(for instanceω) forwhichλ pe k(β). e is called the exponent ofω, equivalently of k(ω).We define the exponent of an algebraic elementαover k to be theinteger e≥o such thatα pe is separable but notα pe−1 . Hence11) Exponent ofα is zero⇔αis separable over k.We shall now extend this notion of exponent and reduced degree toany finite extension.Let K/k be finite so that K= k(ω 1 ,...,ω n ). Put as before K o = k,K i = k(ω 1 ,...,ω i ) so that K n = K. Letω i have reduced degree d i andexponent e i over K i−1 . Then(K i : K i−1 )=d i p e i

4. Separability 29From the definition of d i , it follows that the number of distinct k-isomorphisms of K/k is d 1 ...d n . We putd=d 1···d nand call it the reduced degree of K/k. Then(K : k)=d.... p fwhere f= e 1 +···+e n . We call p f the degree of inseparability of K/k.In order to be able to give another interpretation to the integer d we 34make the following considerations.LetΩ⊃K⊃ k and let K/k have the property that every k automorphismofΩ/k acts like identity on K. Thus ifσ∈G(Ω/k) andω∈K,thenσω=ωAll elements of k have this property. Letω∈K,ωk. Then bydefinition,ω has inΩonly one conjugate. The irreducible polynomialofωhas all roots equal. Thus the minimum polynomial ofωisx pm − awhere a∈k. i.e.,ω pm ∈ k. On the other hand let K be an extension of kinΩwith the property that for everyω∈Kω pm ∈ kfor some integer m ≥ o. Letσbe an automorphism ofΩ/k. Thenσω pm =ω pm . Buto=σω pm −ω pm = (σω−ω) pmwhich shows thatσω=ω. ω being arbitrary in K, it follows that everyelement of G(Ω/k) is identity on K.Hence forω∈Ωthe following three statements are equivalent1) σω=ω for allσ∈G(Ω/k)

30 2. Algebraic extension fields2) ω pm ∈ k for some m≥o depending onω3) The irreducible polynomial ofωover k is of the form x pm − a, a∈k.We call an elementω ∈ Ω which satisfies any one of the above 35conditions, a purely inseparable algebraic element over k.Let us make theDefinition . A subfield K/k ofΩ/k is said to be purely inseparable ifelementω of K is purely inseparable.From what we have seen above, it follows that K/k is purely inseparableis equivalent to the fact that every k-automorphism ofΩis identityon K.Let K/k be an algebraic extension and L the maximal separable subfieldof K/k. For everyω∈K,ω pe is separable for some e≥o whichmeans thatω pe ∈ L. Thus K/L is a purely inseparable extension field.This means that every k-isomorphism of L/k can be extended uniquelyto a k-isomorphism of K/k.Let, in particular, K/k be a finite extension and L the maximal separablesubfield of K. Then K/L is purely inseparable and K/L has noL-isomorphism other than the identity. Thus the number of distinct isomorphismother than the identity. Thus the number of distinct isomorphismsof K/k equals (L : k). But from what has gone before(K : L)= p f , (L : k)=d.For this reason we shall call d also the degree of separability of K/kand denote it by [K : k]. We shall denote the degree of inseparability,p f , by { K : k } . Then(K : k)=[K : k] { K : k } .Note. In case k has characteristic zero, every algebraic element over k isseparable.36IfΩis the algebraic closure of k and L the maximal separable subfieldofΩthenΩ/L is purely inseparable.Ω coincides with L in case k

5. Perfect fields 31has characteristic zero. But it can happen that L is a proper subfield ofΩ.Let K/k be an algebraic extension and L the maximal separable subfield.Consider the exponents of all elements in K. Let e be the maximumof these if it exists. we call e the exponent of the extension K/k. Itcan happen that e is finite but K/L is infinite.If K/k is a finite extension then K/L has degree p f so that the maximume of the exponents of elements of K exists. If e is the exponent ofK/k thene≤ fIt can happen that e< f . For instance let k have characteristic p0and letα∈k be not a p th power in k. Then k(α 1/p ) is of degree p overk. Letβin k be not a p th power in k. Then k(α 1/p ,β 1/p ) is of degree p 2over k,βk(α 1/p ) and for everyλ∈k(α 1/p ,β 1/p ),λ p ∈ k.We may for instance take k(x, y) to be the field of rational functionsof two variables and K = k(x 1/p , y 1/p ). Then (K : k(x, y))= p 2 andλ p ∈ k(x, y) for everyλ∈K.5 Perfect fieldsLet k be a field of characteristic p>0. LetΩbe its algebraic closure.Letω∈k. Then there is only one elementω ′ ∈Ω such thatω ′p =ω.We can therefore writeω 1/p without any ambiguity. Let k p−1 be the fieldgenerated inΩ/k by the p th roots of all elements of k. Similarly from 37k p−2 ,... Let ⋃K= k p−nn≥0Obviously K is a field; for ifα,β∈K,α,β∈k p−n for some large n.We denote K by k p−∞We shall study k p−∞ in relation to k andΩ. k p−∞ is called the rootfield of k.Letω∈k p−∞ . Thenω∈k p−∞ for some n so thatω p∞ ∈ k orωispurely inseparable. On the other hand letω∈Ω be purely inseparable.Thenω pn ∈ k for some n i.e.,ω∈k p−∞ ⊂ k p−∞ . Thus

32 2. Algebraic extension fields1) k p−∞ is the largest purely inseparable subfield of Ω/k.Therefore every automorphism ofΩ/k is identity on k p−∞ . The setof elements ofΩwhich are fixed under all the k− automorphisms ofΩ/k form a field called the fixed of G(Ω/k). Since every such elementω,fixed under G(Ω/k) is purely inseparable,Ω∈k p−∞ . Hence2) k p−∞ is the fixed field of the group of k− automorphism ofΩ/k.Let f (x) be an irreducible polynomial in k p−∞ [x]. We assert that thisis separable. For if not f (x)∈k p−∞ [x p ]. Thus f (x)=a o + a 1 x p +···+a n x np . Since a i ∈ k p−∞ [X p ], it is in some k p−t and so a i = b p iforb i ∈ k p−∞ . Hencef (x)=(b o + b 1 x+···+b n x n ) pwhich is contradicts the fact that f (x) is irreducible. Hence383) Ω/k p−∞ is a separable extension.We now make theDefinition. A field k is said to be perfect if every algebraic extensionof k is separable.It follows from the definition that1) An algebraically closed field is perfect2) A field of characteristic zero is perfect.We shall now prove3) A field k of characteristic p>0 is perfect if and only if k=k p−∞ .Let k have no inseparable extension. Then for a∈k, a 1/p ∈ kalso; for, otherwise k(a 1/p ) is inseparable over k. Thus k=k p−1 =···=k p−∞ . The converse has already been proved.We deduce immediately4) A finite field is perfect.For if k is a finite field of characteristic p>rthen a→a p is anautomorphism of k.

5. Perfect fields 335) Any algebraic extension of a perfect field is perfect.For let K/k be algebraic and k be perfect. Ifαis inseparable overK, then it is already so over k.An example of an imperfect field is the field of rational functionsof one variable x over a finite field k. For if k has characteristicp, then x 1/p k(x) and k(x 1/p ) is a purely inseparable extensionover k(x).Note 1. Ifα∈Ω is inseparable over k, it is not true that it is inseparableover every intermediary field, whereas this is true ifαis separable. 39Note 2. If K/k is algebraic and K∩k p−∞ contains k properly then K is aninseparable extension. But the converse of this is not true, that is, if K/kis an inseparable extension, it can happen that there are no elements inK which are purely inseparable over k. We give to this end the followingexample due to Bourbaki.Let k be a field of characteristic p>2 and let f (x) by in irreduciblepolynomialf (x)= x n + a 1 x n−1 +···+a nin k[x]. Ifα 1 ,...,α t are the distinct roots of f (x) inΩthenf (x)= { (x−α 1 )...(x−α t ) } p e e≥1.where n=t.· p e . Putφ(x)= f (x p ). ThenIfβ i =α 1/pithenφ(x)= { (x p −α 1 )...(x p −α t )} peφ(x)= { (x−β 1 )...(x−β t ) } p e+1andβ,...,β t are distinct sinceα 1 ...α t are distinct. Supposeφ(x) isreducible in k[x] and letψ(x) be an irreducible factor ofφ(x) in k[x].Thenψ(x)= { (x−β 1 )···(x−β t ) } p µ

34 2. Algebraic extension fieldsforµ≥0 andl≤t. (This is because roots ofψ(x) occur with the samemultiplicity). We can writeψ(x)= { (x p −α 1 )...(x p −α l ) } p µ−1(µ has to be≥ 1) since otherwise, it will mean that (x−α 1 )···(x−α t )∈k[x]). Now this will mean thatφ(x)=ψ(x). W(x) in k[x p ] so thatl=t.Henceψ(x)={(x−β 1 )···(x−β t )} pµ ,µ≥1.40Sinceψ(x) is irreducible andψ(x)={(x p −α 1 )...(x p −α t )} pµ−1we see thatµ−1=eorµ=e+1. Thusφ(x)= f (x p )={ψ(x)} pThus if f (x p ) is reducible, it is the p th power of an irreducible polynomial.In this case a i = b p i , b i∈ k, i=1,...n.Conversely if a i = b p i , b i∈ k, i=1,...n. then f (x p ) is reducible.Hence f (x p ) is reducible f (x)∈k p [x].Let k now be a field of characteristic p>2given by k=Γ(x, y), thefield of rational functions in two variables x, y over the prime fieldΓifp elements. Consider the polynomialf (z)=z 2p + xz p + y,in k[z]. Since x 1/p , y 1/p do not lie in k, f (z) is irreducible in k. Letϑbea root of f (z). Then (k(ϑ) : k)=2p. Letβbe in k(ϑ) and not in k suchthatβ p ∈ k. Then k(ϑ)⊃k(β)⊃k. Also (k(β) : k)= p. In k(β)[x]the polynomial f (z) cannot be irreducible, since ( k(ϑ) : k(β) ) = 2. It isreducible and so k(β) will contain x 1/p and y 1/p . But then k(x 1/p , y 1/p )⊃k(β).Thusp 2 = ( k(x 1/p , y 1/p ) : k ) ≤ ( k(β) : k ) ≤ ( ϑ) : k ) = 2pbut this is impossible. Thus there is no elementβin k(ϑ) withβ p ∈ k.All the same k(ϑ) is inseparable.

6. Simple extensions 356) If k is not perfect then k p−∞ is an infinite extension of k.For, if k p−∞ /k is finite then, since k p−∞ = ⋃ k p−n we have k p−n =nk p−(n+1) for some n,Applying the mapping a→a apn we find k=k p−1 . But this is false. 41Thus (k p−(n+1) : k p−n) > 1which proves our contention.6 Simple extensionsAn algebraic extension K/k is said to be simple if there is anω∈Ksuchthat K= k(ω). Obviously (K : k) is finite. We callωaprimitive elementof K. The primitive element is not unique for,ω+λ,λ ∈ k also isprimitive. We now wish to find conditions when an algebraic extensionwould be simple. We first proveLemma. Let k be an infinite field andα,β elements in an algebraic closureΩofk such thatαis separable over k. Then k(α,β) is a simpleextension of k.Proof. Let f (x) andφ(x) be the irreducible polynomials ofαandβrespectivelyin k[x], so thatf (x)=(x−α 1 )···(x−α n )φ(x)=(x−β 1 )···(x−β m ).Sinceαis separable,α 1 ,...α n are all distinct. We shall putα=α 1 ,andβ=β 1 . Construct the linear polynomialsβ i + Xα j (i=1,...,m; j=1,...,n)These mn polynomials are inΩ, the algebraic closure of k and sincek is infinite, there exists an elementλ∈ksuch thatβ i +λα j β i ,+λα j ,α J ′□

36 2. Algebraic extension fields42Putγ=β 1 +λα 1 =β+λαObviouslyλcan be chosen so thatγ0Now k(γ)⊂k(α,β). We shall will now show thatα∈k(γ). Fromdefinition ofγ,βalso will be in k(γ) and that will meank(γ)⊂k(α,β)⊂k(γ).In order to do this consider the polynomialφ(γ−λ.x)in k(γ)[x]. Alsoit vanishes for x=α. Furthermore by our choice ofλ,φ(γ−λα i )0for i>1. In the algebraic closureΩ, therefore, f (x) andφ(γ−λx) havejust x−α as a factor. But f (x) andφ(γ−λx) are both polynomials ink(γ)[x]. So x−αis the greatest common divisor ofφ(γ−λx) and f (x)in k(γ)[x]. Thusα∈k(γ). Our lemma is demonstrated.We have thereforeCorollary . Ifα 1 ,...,α n are separably algebraic and β ∈ Ω thenk(β,α,...α n ) is a simple extension.We deduce immediatelyCorollary. A finite separable extension is simple.LetΓbe the field of rational numbers andΓ(ω,ρ) the splitting fieldof the polynomial x 3 − 2 inΓ[x]. ThenΓ(ω,ρ) is simple. A primitiveelementγis given byω+ρ=γ. ThenΓ(γ) is of degree 6 overΓ. It iseasy to see thatγhas overΓthe minimum polynomial(x 3 − 3x−3) 2 + 3x(x+1)(x 3 − 3x−3)+9x 2 (x+1) 2Let now K/k be a finite extension and L the maximal separablesubfield of K/k. Then (K : L)= p f the degree of inseparability and(L : k)=d the reduced degree. If we consider the exponents of elementsof K, these have a maximum e ande≤ f.43We had given an example of e< f . We shall now prove the

6. Simple extensions 37Theorem 5. If e is the exponent and p f the degree of inseparability offinite extension K of k, then e= f⇐⇒ K/k is simple.Proof. Let K= k(ω). Let K o be the maximal separable subfield of K/k.Now (K : K o )= p f . But p f is degree ofω. Thus e= f □Let now K/k be a finite extension and e= f . There exists then aωin K such thatω pe is separable and in K o but for no, t

38 2. Algebraic extension fieldsThereforek(α,β)⊂k(γ)⊂k(α,β).Hence every subfield of K, generated by 2 and hence by a finite numberof elements is simple. Let K be a maximal subfield of K/k whichis simple. (This exists since K/k has only finitely many intermediaryfields). Let K 0 = k(ω). Letβ∈KandβK 0 . Then k(γ)=k(ω,β)⊂Kand K 0 ⊂ k(γ) contradicting maximality of K 0 . Thusβ(γ)=K. We haveprovedTheorem 6. K/k is simple⇐⇒ K/k has only finitely many intermediaryfields.We deduceCorollary. If K/k is simple, then every intermediary field is simple.Note 1. We have the fact that if K/k is infinite there exist infinitely manyintermediary fields.45Note 2. Theorem 6 has been proved on the assumption that k is an infinitefield. If k is finite the theorem is still true and we give a prooflater.7 Galois extensionsLet K/k be an algebraic extension and G the group of automorphism ofK which are trivial on k. Let L be the subset of all elements of K whichare fixed by G. L is then a subfield of K and is called the fixed field ofG. We shall now consider the class of algebraic extensions K/k whichare such that the group G(K/k) of automorphisms of K which are trivialon k, has k as the fixed field. We call such extensions galois extensions,the group G(K/k) itself being called the galois group of K/k.We now prove theTheorem 7. k is the fixed field of the group of k automorphisms ofK⇐⇒ K/k is a normal and separable extension.

7. Galois extensions 39Proof. Let k be the fixed field of the group G(K/k) of k-automorphismsof K. Letω∈K. Letω 1 (=ω),...ω n be all the distinct conjugates ofωthat lie in K. Consider the polynomialf (x)=(x−ω 1 )···(x−ω n )Ifσis an element of G(K/k),σ permutesω 1 ,...,ω n so thatσleavesthe polynomial f (x) unaltered. The coefficients of this polynomial arefixed under all elements of G and hence since k is the fixed field G, f (x)∈k[x]. Hence the minimum polynomial roots of the minimum polynomialofω, sinceω 1 ,...ω n are conjugates. Thus f (x)/φ(x). ThereforeK splitting field ofφ(x),φ(x) has all roots distinct. Thus K/k is normaland separable.□Suppose now K/k is normal and separable. Consider the group 46G(K/k) of k-automorphisms of K. Letα∈K. Since K/k is separable,all conjugates ofαare distinct. Also since K/k is normal K contains allthe conjugates. Ifαis fixed under allσ∈G(K/k), thenαis a purelyinseparable element of K and hence is in k.Our theorem is thus proved.We thus see that galois extensions are identical with extension fieldswhich are both normal and separable.Examples of Galois extensions are the splitting fields of polynomialsover perfect fields.Let k be a field of characteristic 2 and let K= k( √ α) forα∈kand √ αk.( √ α) 2 =α∈k. Every element of K is uniquely of the forma+ √ α·b, a, b∈k. Ifσis an automorphism of K which is trivial on k,then its effect on K is determined by its effect on √ α. Nowα=σ { ( √ α) 2} =σ( √ α).σ( √ α)or thatσ( √ α)/ √ α=λ is such thatλ 2 = 1. Sinceλ∈K,λ=±1. Thusσ is either th identity or the automorphismσ( √ α)=− √ αThus G(K/k) is a group of order 2. K/k is normal and separable.We shall obtain some important properties of galois extensions.

40 2. Algebraic extension fields471) If K/k is a galois extension and k⊂L⊂K, then K/L is a galoisextension also.For, K/L is clearly separable. We had already seen that it is normal.If we denote by G(K/L) the galois group of K over L, than G(K/L)is a subgroup of G(K/k).2) If k⊂L 1 ⊂ L 2 ⊂ K then G(K/L 2 ) is a subgroup of G(K/L 1 ). Thisis trivial.3) If{K α } is a family of galois extensions of k contained inΩthen ⋂ αand k( ⋃ K α ) are galois.αK αThis follows from the fact that this is already true for normal andalso for separable extensions.4) If K/k is galois and L and L ′ are two intermediary fields of K/kwhich are conjugate over k, then G(G/L) and G(K/L ′ ) are conjugatesubgroups of G(K/k) and conversely.Proof. Since L and L ′ are conjugate over k letσbe an automorphism ofK/k so thatσL= L ′ . Letτ∈G(K/σL). Then for everyω∈σLButω=σω ′ forω ′ ∈ L. Thusτω=ωσ −1 τσω ′ =ω ′Since this is true for everyω ′ ∈ L, it follows thatσ −1 G(K/σL)σ⊂G(K/L)In a similar manner one proves thatσG(K/L)σ −1 ⊂ G(K/σL) whichproves our contention. □48Conversely suppose that L and L ′ are two subfields such that G(K/L)and G(K/L ′ ) are conjugate subgroups of G(K/k). Let G(K/L ′ )=σ −1G(K/L)σ. Letω∈L ′ andτ∈G(K/L). Thenσ −1 τσ∈G(K/L ′ ) and so

7. Galois extensions 41σ −1 τσω=ωorτ(σω)=σω. This being true for allτ, it follows thatσω∈L for allω in L ′ . ThusσL ′ ⊂ L ′ . We can similarly prove thatσ −1 L⊂L ′ whichproves our statement.In particular let L/k be a normal extension of k which is containedin K. ThenσL=L for allσ∈G(K/k). This means that G(K/L) is anormal subgroup of G(K/k). On the other hand if L is any subfield suchthat G(K/L) is a normal subgroup of G(K/k) then by aboveσL=L forallσ∈G(K/k) which proves that L/k is normal. Thus5) Let k⊂L⊂K. Then L/k is normal⇐⇒ G(K/L) is a normalsubgroup of G(K/k)6) If L/k is normal, then G(K/k)≃G(L/k)/G(K/L).Letσ∈G(K/k) and ¯σ the restriction ofσto L. Then ¯σ is an automorphismof L/k so that ¯σ∈G(L/k). Nowσ→ ¯σ is a homomorphismfo G(K/k) into G(L/k). Forστω=στω=σ(τω)= ¯σ¯τω for allω∈L. Thus ¯σ¯τ=στThe homomorphism is onto since every automorphism of L/k can be 49extended into an automorphism of K/k. Now ¯σ is identity if and only if¯σω=ωfor allω∈L. Thusσ∈G(K/L). Also everyσ∈G(K/L) has thisproperty so that the kernel of the homomorphism is G(K/L).Let K/k be a finite galois extension. Every isomorphism of K/k inΩ is an an automorphism. Also K/k being separable, K/k has exactly(K : k) district isomorphisms. This shows that(K : k)=order of G.We shall now prove the converse7) If G is a finite group of automorphisms of K/k having k as thefixed field then (K : k)=order of G.

42 2. Algebraic extension fieldsProof. Letσ 1 ,...,σ n be the n elements of G andω 1 ,...ω n+1 any n+1elements of K. Denote by V K the vector space over K of n dimensionsformed by n- tuples (α 1 ,...,α n ). Define n+1 vectorsΩ 1 ,...,Ω n+1 byΩ i = (σ i (ω i ),...,σ n (ω i ))i=1,...,n+1Among these vectors there exists m≤n vectors linearly independentover K. LetΩ 1 ,...,Ω m be independent. Thenm∑Ω m+1 = a i Ω in=1a i ∈ KThis equation gives, for the components of theΩ ′ s,σ l (ω m+1 )=m∑a i σ l (ω i )α=1,...,n.1=i□50Sinceσ 1 ,...,σ n form a group thenσ n σ 1 ,...,σ n σ n are again theelementsσ 1 ,...,σ n in some order. Thusσ h σ l (ω m+1 )=m∑σ h (a i |σ h σ l (ω i )i=1which meansσ l (ω m+1 )=m∑σ h (a i )σ l (ω i )i=1i=1,...,nsubtracting we havem∑(σ h (a i )−a i σ l (ω i )=0i=1which means thatm∑ ( )σh (a i )−a i Ωi = 0i=1

7. Galois extensions 43From linear independence, it follows thatσ h (a i )=a i for all i. Buth is arbitrary. Thus a i ∈ k. We therefore have by takingσ 1 to be theidentity element of Gm∑ω m+1 = a i ω ia i are in k, not all zero. Hencei=1(K : k)≤n.But every element of G is an isomorphism of K/k. Thus(K : k)≥ order of G=n.Our assertion is established.Suppose K/k is a galois extension. For every subfield L of K/k, theextension K/L is galois. We denote its galois group by G(L) and thisis a subgroup of G(K/k). Suppose g is any subgroup of G(K/k) and letF(g) be its fixed field. Then F(g) is a subfield of K. The galois groupG(F(g)) of K/F(g) contains g. In general one has only 51g⊂G(F(g))Let now K/k be a finite galois extension. Let g be a subgroup ofG(K/k)=Gand F(g), the fixed field of g. Then by above( )K : F(g) = order of g.and sog=G ( K/F(g) )If g 1 and g 2 are two subgroups of G with g 1 ⊂ g 2 then F(g 1 ) andF(g 2 ) are distinct. For if F(g 1 ) and F(g 2 ) are identical, then by aboveg 1 = G(K/F(g 2 ))=g 2 . We thus have theMain Theorem(of finite galois theory). Let K/k be a finite galois extensionwith galois group G. Let M denote the class of all subgroupsof G and N the class all subfield of K/k. Letφbe the mapping which

44 2. Algebraic extension fields52assigns to every subgroup g∈ M, the fixed field F(g) of g in N. Thenφis a mapping of M onto N which is biunivocal.In order to restore this property even for infinite extensions, we developa method due originally to Krull.Let K/k be a galois extension with galois group G(K/k). For everyω ∈ K, we denote by G ω the galois group G(K/k(ω)). This then isa subgroup of G(K/k). We make G(K/k) into a topological group byprescribing the{G ω } as a fundamental system of neighbourhoods of theidentity element e∈G(K/k). Obviously ⋂ αG α = (e). Forσ∈ ⋂ α G α ⇒σα=αfor allα∈ K so thatσ=e. It is easy to verify that{G α } satisfythe axioms for a fundamental system of open sets containing the identityelements e.Any open set in G is therefore a union of sets of type{σG α } or afinite intersection of such. Also since G α are open subgroups they areclosed; for,σG α is open for allσand hence⋃σG ασeis also open. Therefore G α is closed. This proves that the topology onG makes it totally disconnected. We call the topology on G the Krulltopology.If g is a subgroup of G, ḡ the closure of g is also a subgroup. Wenow prove theLemma. Let g be a subgroup of G a and L its fixed field. ThenG(K/L)=ḡ(the closure of g).Proof. Letωbe an element in K and f (x) its minimum polynomial ink. Consider f (x) as a polynomial over L and let L ′ be its splitting fieldover L. Then L ′ /L is a galois extension. The restriction of elements ofg to L ′ are automorphisms of L ′ with L as fixed field (by definition ofL). By finite galois theory these are all the elements of the galois groupof L ′ /L. This means that every automorphism of L ′ /L comes from anelements of g.□

7. Galois extensions 45Letσ ∈ G(K/L). Letωbe any element in K and G the groupG(K/k(ω)). The restriction ofσto L ′ is an automorphism of L ′ withL as fixed field. There is thus an elementτ∈g, which has on L ′ thesame effect asσ. Henceτ −1 σ is identity on L ′ and sinceω∈L ′ we getτ −1 σω=ωThis means thatτ −1 σ∈G ω by definition of G ω . Henceσ −1 τ∈G ω 53orτ∈σG ω . ButσG ω is an open set containingσ. Therefore since{σG ω } for all G ω form a fundamental system of neighbourhoods ofσwe conclude thatσ∈ḡ.Thus G(K/L)⊂ḡLet nowσ∈ḡ. ThenσG ω is a neighbourhood ofσand so intersectsg in a non empty set. Letω∈L. Letτ∈σG ω ∩ g. Letσ ′ in G ω suchthatσσ ′ =τBy definition ofτ,τω=ω. Butτω=σσ ′ ω=ω. Thereforeσσ ′ ∈ G ω .Butσ ′ is already is in G ω . Thereforeσω=ω. Alsoωbeing arbitraryσL=Lwhich means thatσ∈G(K/L). Thusḡ⊂G(K/L)and our contention is established.From the lemma, it follows that if L is an intermediary field, thegalois group G(K/L) is a closed subgroup of G(K/k). On the other handif g is a closed subgroup of G and F(g) its fixed field thenwe have hence the fundamentalG ( K/F(g) ) = ḡ=g.Theorem 8. Let K/k be a galois extension and G(K/k) the galois groupwith the Krull - topology. Let M denote the set of closed subgroups ofG and N the set of intermediary fields of K/k. Letφbe the mappingwhich assigns to every g∈ M, the fixed field F(g) of g in N. Thenφis a 54biunivocal mapping of M on N.

46 2. Algebraic extension fields55Suppose now that K/k is a galois extension and L is an intermediaryfield. Let G(K/k) and G(K/L) be the galois groups. On G(K/L) thereare two topologies, one that is induced by the topology on G(K/k) andthe other the topology that G(K/L) possesses as a galois extension. IfL ′ is a subfield of K/L so that L ′ /L is finite, then G(K/L ′ ) is an open setin the inherent topology on G(K/L). On the other hand L ′ = L(ω) sinceL ′ /L is a separable extension.ThusG(K/L ′ )⊂G ω ∩ G(K/L)which proves that the two topologies are equivalent. Here we have usedthe fact that a finite separable extension of L is simple. We gave alreadyseen the truth of this statement if L is infinite. In case L is finite it isproved in the next section. ‘ In a similar manner if K/k is galois and L isa normal extension of k in K, then on G(L/k) there are two topologies,one the inherent one and the other the topology of the quotient groupG(K/k)/G(K/L). One can prove that the two topologies are equivalent.We call an extension K/k abelian or solvable according as G(K/k)is abelian or a solvable group. If K/k is a galois extension and G(K/k)its galois group with the Krull topology let H denote the closure of thealgebraic commutator subgroup of G(K/k). H is called the topologicalcommutator subgroup. If L is its fixed field, then since H is normal, L/kis a galois extension. Its galois group is isomorphic to G/H which isabelian. From the property of the commutator subgroup, it follows thatL is the maximal abelian subfield of K/k.8 Finite fieldsLet K be a finite field of q elements, q= p n where p is the characteristicof K. LetΓbe the prime field of p elements. Then(K :Γ)=nK ∗ the group of non-zero elements of K is an abelian group of orderq−1. Forα∈K ∗ we haveα q−1 = 1

8. Finite fields 471 being the unit element of K. The q−1 elements of K ∗ are roots ofx q−1 − 1. Also ∏x q−1 − 1= (x−α).α∈K ∗Letα∈K ∗ . Let d be its order as an element of the finite group K ∗ .Thenα d = 1. Consider the polynomial x d − 1. It has in K ∗ at most droots. Also d/q−1. Butx q−1 − 1=(x d − 1)(x q−1−d +···)Since x d − 1 and x q−1−d +··· both have respectively at most d andq−1−d roots in K ∗ and they together q−1 roots in K ∗ it follows thatfor every divisor d of q−1, x d − 1 has exactly d roots in K ∗ . These rootsfrom a group of order d. If it is cyclic then there is an element of orderd and there are exactlyφ(d) elements of order d. Also∑φ(d)=q−1d/q−1which proves that for every divisor d of q−1 there areφ(d)≥1 elements 56of order d. Thus1) The multiplicative group of a finite field is cyclic.Let k be a finite of q elements and K a finite extension of k of degreen. Then K has q n elements. Since K ∗ is cyclic, letρbe a generatorof K ∗ . ThenK= k(ρ)which proves2) Every finite extension of a finite field is simple.For a finite field of characteristic p, a→a pe is an automorphism ofK. Since k has q elements we havefor every a∈k.a q = a

48 2. Algebraic extension fieldsNow a→a q is an automorphism of K/k which fixes elements of k.Call this automorphismσ.σ is determined uniquely by its effect ona generatorρ of K ∗ . Consider the automorphismsThese are distinct. For,1,σ,σ 2 ,...,σ n−1σ i ρ=σ i−1 (σρ)=σ i−1 (ρ q )=ρ qiHenceρ q = 1⇐⇒ q i ≡ o( mod q n ) or i=o (Since i

Chapter 3Algebraic function fields1 F.K. Schmidt’s theoremLet K/k be an extension field and x 1 ,..., x n+1 any n+1 elements of 58K. Let R=k[z 1 ,...,z n+1 ] be the ring of polynomials in n+1 variablesover k. Let Y be the ideal in R of polynomials f (z 1 ,...,z n+1 ) with thepropertyf (x 1 ,..., x n+1 )=0.Then clearly Y is a prime ideal of R. Also since R is a Noetheeianring, Y is finitely generated. Note that Y = (o) if and only if x 1 ,...,x n+1 are algebraically independent over k. Y is called the ideal of theset x 1 ,..., x n+1 .We shall consider the case where the set x 1 ,..., x n+1 has dimensionn over k, that is that k(x 1 ,..., x n+1 ) is of transcendence degree n over k.We proveTheorem 1. Y is a principal ideal generated by an irreducible polynomial.Proof. Without loss in generality we may assume that x 1 ,..., x n are algebraicallyindependent over k and that x n+1 is algebraic overk(x 1 ,..., x n ), so that Y (o). □49

50 3. Algebraic function fieldsConsider the degrees of the polynomials f (z 1 ,...,z n+1 ) (in Y ) inthe variable z n+1 . These degrees have a minimum greater than zero sinceY (o), and x 1 ,..., x n are algebraically independent over k. Letϕbea polynomial in Y of smallest degree in z n+1 . Putϕ=A o z λ n+1 + A 1z λ−1n+...+A λ59where A 0 , A 1 ,..., A λ are polynomials in z 1 ,...z n with coefficients ink. We may assume that A 0 , A 1 ,..., A λ have no common factor ink[z 1 ,...,z n ]. For if A(z 1 ,...,z n ) is a common factor of A 0 ,..., A λ thenϕ(z 1 ,...,z n+1 )=A(z 1 ,...,z n )ϕ 1 (z 1 ,...,z n+1 )and so, since x 1 ,..., x n are algebraically independent,ϕ 1 (x 1 ,..., x n+1 )=oandϕ 1 will serve our purpose. So we can takeϕto be a primitive polynomialin z n+1 over R ′ = k[z 1 ,...,z n ].Clearlyϕis irreducible in R ′ . For, ifϕ=g 1 (z 1 ,...,z n+1 ) g 2 (z 1 ,...,z n+1 )then g i (x 1 ,..., x n+1 )=0 for i=1 or 2, so that either g 1 or g 2 is in Y .Both cannot have a term in z n+1 with non zero coefficient. For then thedegrees in z n+1 of g 1 of g 2 will both be less that ofϕin z n+1 contradictingthe definition ofφ. So one g 1 , g 2 say g 1 is independent of z n+1 . But thismeans thatϕis not a primitive polynomial.Thus we have chosen in Y a polynomialϕwhich os irreducible, ofthe smallest degree in z n+1 and primitive in R ′ [z n+1 ].Letψ(z 1 ,...,z n+1 ) be any other polynomial in Y .Since F= k(z 1 ,...,z n ) is a field, F[z n+1 ] is a Euclidean ring so thatin F[z n+1 ] we haveψ(z 1 ,...,z n+1 )=A(z 1 ,...,z n+1 )ϕ(z 1 ,...,z n+1 )+ L(z 1 ,...,z n+1 )60where A and L are polynomials in z n+1 over F. Here either L=o ordegree of L in z n+1 is less than that ofϕ. If Lo, then we may multiply

1. F.K. Schmidt’s theorem 51both sides of the above equation by a suitable polynomial in z 1 ,...,z nover k so thatB(z 1 ,...,z n )ψ= C(z 1 ,...,z n+1 )ϕ(z 1 ,...,z n+1 )+ L 1 (z 1 ,...,z n+1 )L 1 having in z n+1 the same degree as L. Sinceϕandψare in Y , itfollows that L 1 ∈ Y . Because of degree of L 1 , it follows thatThusL 1 = L=0.B(z 1 ,...,z n )ψ=A(z 1 ,...,z n+1 )ϕ(z 1 ,...,z n+1 )Sinceϕis a primitive polynomial, it follows thatϕdividesψand ourtheorem is proved.We callϕthe irreducible polynomial of x 1 ,... x n+1 over k.Note that since x 1 ,... x n are algebraically independent over k, thepolynomialϕ 1 (z n+1 )=ϕ(x 1 ,..., x n , z n+1 )over k(x 1 ,..., x n ) is irreducible in z n+1 .Let x 1 ,..., x n+1 be of dimension n over k andϕthe irreducible polynomialof x 1 ,..., x n+1 over k. Letϕbe a polynomial in z 1 ,...,z i+1 butnot in z i+2 ,...,z n+1 , that is it does not involve z i+2 ,...,z n+1 in its expression.Consider the field L=k(x 1 ,..., x i+1 ). because x 1 ,..., x i+1 arealgebraically dependent (ϕ(x 1 ,..., x i+1 )=0), 61dim k L≤i.But k(x 1 ,..., x n+1 )=L(x i+2 ,..., x n+1 ) so thatdim L k(x 1 ,..., x n+1 )≤n−i.Since dimensions are additive, we haven=dim k L+dim L k(x 1 ,..., x n+1 )≤i+n−i=n.Thus L has over k the dimension i. Sinceϕis a polynomial inz 1 ,...,z i+1 every one of these variables occurring, with non zero coefficients,we get the

52 3. Algebraic function fieldsCorollary. If x 1 ,..., x n+1 be n+1 elements of K/k and have dimensionn, there exist among them i+1 elements, i≤n, say x 1 ,..., x i+1 (in someorder) such that k(x 1 ,... x i+1 ) has dimension i over k and every i of themare algebraically independent.Let K/k be a transcendental extension with a transcendence base Bover k. Then K/k(B) is algebraic. We call K/k an algebraic functionfield if(1) B is a finite set(2) K/k(B) is finite algebraic.62Let dim k K = n. There exist x 1 ,..., x n in K which form a transcendencebase of K/k. If K is an algebraic function field then K/k(x 1 ,..., x n ) is finite algebraic. Hence K= k(x 1 ,..., x m ), m≥n, is finitelygenerated. This shows that algebraic function fields are identical withfinitely generated extensions.An algebraic function field K/k is said to be separably generatedif there exists a transcendence base x 1 ,..., x n of K/k such thatK/k(x 1 ,..., x n ) is a separable algebraic extension of finite degree.x 1 ,..., x n is then said to be a separating base. Clearly every purely transcendentalextension is separably generated. Also, if k has characteristiczero and K is an algebraic function field, it is separably generated. Inthis case every transcendence base is a separating base. This is no longertrue if k has characteristic po.For example, let K = k(x, y) be a function field of transcendencedegree one and letx 2 − y p = o.Let k have characteristic p2. Obviously x and y are both transcendentalover k. But K/k(x) is a simple extension generated by y which isa root ofz p − x 2over k(x)[z] and since k has characteristic p, K/k(x) is purely inseparable.On the other hand K/k(y) is separable since x satisfies over k(y) thepolynomialx 2 − y p .

1. F.K. Schmidt’s theorem 53Thus y is separating but not x.An algebraic function field which is not separably generated is saidto be inseparably generated. This means that for every base B of K/k.K/k(B) is inseparably algebraic.In algebraic geometry and in algebraic function theory, it is of im- 63portance to know when an algebraic function field is separably generated.An important theorem in this regard is theorem 2 due to F.K.Schmidt. We shall first prove aLemma . Let k be a perfect field of characteristic p o and K =k(x 1 ,..., x n+1 ) an extension field of dimension n. Then K is separablygenerated.Proof. Letϕbe the irreducible polynomial of x 1 ,..., x n+1 and let it bea polynomial in z 1 ,...,z i+1 but in z i+2 ,...,z n+1 . Thenϕ(x 1 ,...,z t ,..., x i+1 )is irreducible over k(x 1 ,..., x t−1 , x t+1 ,..., x i+1 ) for every t, 1≤t≤i+1.At least for one t,ϕ(z 1 ,...,z t ,...,z i+1 ) is a separable polynomial in z tover k(z 1 ,...,z t−1 ,...,z i+1 ). For, if it is inseparable in every z t , thenϕ(z 1 ,...,z i+1 )∈k[z p 1 ,...,zp i+1 ]and since k is perfect, this will mean thatϕ(z 1 ,...,z i+1 ) is the pth powerof a polynomial in k[z 1 ,...,z i+1 ] which contradicts irreducibility ofϕ.So, for some z t , say z 1 , we haveϕ(z 1 , x 2 ,..., x i+1 ) is a separable polynomial.Hence x 1 is separable over k(x 2 ,..., x i+1 ) and so over k(x 2 ,..., x n+1 ). But x 2 ,..., x n+1 has dimension n and our lemma is proved.□Corollary. Under the conditions of the lemma, a separating base of nelements may be chosen from among x 1 ,..., x n+1 .We are now ready to prove the theorem of F.K.Schmidt. 64Theorem 2. Every algebraic function field K over a perfect field k isseparably generated.

54 3. Algebraic function fieldsProof. Obviously, the theorem is true if k has characteristic zero. Solet k have characteristic po. Let K= k(x 1 ,..., x m ) and let n be thedimension of K/k. Then m≥n. If m=n there is nothing to prove.Let m=n+q. If q=1 then lemma 1 proves the theorem. So let usassume theorem proved for q−1 instead of q>1. We may assume,without loss in generality that x 1 ,..., x n is a transcendence base of K/k.Consider the fields L=k(x 1 ,..., x n , x n+1 ). It satisfies the conditionsof the lemma. Hence there exist n elements among x 1 ,..., x n+1 sayx 1 ,..., x t−1 , x t+1 ,... x n+1 which from a separating base of L/k. Thus x tis separable over k(x 1 ,..., x t−1 ,..., x n+1 ) and hence overM= k(x 1 ,..., x t−1 , x t+1 , x t+2 ,..., x m )M/k now satisfies the induction hypothesis and so is separably generated.Since K/M is separable, it follows that K is separably generated.□We could prove even more if we assume as induction hypothesis thefact that among x 1 ,..., x m there exists a separating base.2 DerivationsLet R be a commutative ring with unit element e. A mapping D of Rinto itself is said to be a derivation of R if65(1) D(a+b)=Da+Db(2) D(ab)=aDb+bDafor a, b∈R. It is said to be a derivation over a subring R ′ if for everya∈R ′ , Da=o. It then follows that for a∈R ′ and x∈RDax=aDxThe set R o of a∈R with Da=o is a subring of R and contains e.For,De=De 2 = De. e+e. De=2De ;

2. Derivations 55so De=o. If Da=o, Db=o, thenD(a+b)=Da+Db=oD(ab)=aDb+bDa=oThus R o is a subring. We call R o the ring of constants of the derivationD.If R is a field, then R o also is a field. For, if x∈R o and xo, theno= De=Dx.x −1 = Dx.x −1 + x.Dx −1ThusDx −1 = oso that x −1 ∈ R o .D is said to be a non-trivial derivation of R if there is an x∈R withDxo. It follows from above thatTheorem 3. A prime field has no non-trivial derivations.A Derivation ¯D of R is said to be an extension of a derivation D of asubring R ′ of R if ¯Da=Da for a∈R ′ . We now prove the 66Theorem 4. If K is the quotient field of an integrity domain R, then aderivation D of R can be uniquely extended to K.Proof. Every element c in K can be expressed in the form c= a b , a,b∈R. If an extension ¯D of D exists, thenBut a=bc so that¯Da=DaDa= ¯Da= ¯Dbc=b¯Dc+cDbTherefore¯Dc= Da−cDbb= bDa−aDbb 2

56 3. Algebraic function fieldsor thatIf c is expressed in the form a′b ′ , a′ , b ′ ∈ R then ab ′ = ba ′ and soDa.b ′ + a.Db ′ = Db.a ′ + bDa ′Da ′ − cDb ′b ′= Da−cDbbwhich proves that ¯Dc does not depend on the way c is expressed asthe ratio of two elements from R. We have therefore only to prove theexistence of ¯D. In order to prove this, put for c= a b¯Dc= bDa−aDbb 2 ;we should verify that it is a derivation, is independent of the way c isexpressed as ratio of elements in R and that it coincides with D on R.These are very simple. □67Let D 1 , D 2 be two deviations of R. Define D=D 1 + D 2 by Da=D 1 a+D 2 a for a∈R. Then it is easy to verify that D is a derivation ofR. Furthermore if a∈R define aD by(aD)x=a.DxBy this means, the derivations of R from an R module, SupposeD 1 ,..., D r from a basis of the module of derivations of R.Then every derivation D of R is of the form∑D= a i D i , a i ∈ R.iLet R be an integrity domain and ¯D 1 ,..., ¯D r the unique extensionsof D 1 ,..., D r respectively to K, the quotient, field of R. Then ¯D 1 ,..., ¯D rare linearly independent over K. For, if∑1 i ¯D i = o, 1 i ∈ Ki

2. Derivations 57then write 1 i = a ib i, a i , b i ∈ R. We get∑iaibi ¯D i = o.Multiplying throughout by b 1 ,...,b r which is not zero, we get( ∑ λ ¯D i )t = o for every t ∈ R. Therefore ∑ λ i D i = o which impliesiithatλ i = o or a i = o.Also, since every derivation D of K is an extension of a derivationof R, it follows that the derivations of K from an r-dimensional vectorspace over K.Let us now consider the case where R=k[x 1 ,..., x n ] is the ring ofpolynomials in n variables x 1 ,..., x n . The n mappingsD i :a→ ∂a∂x i, i=1,...,nare clearly derivations of R over k. They form a base of derivations of 68R which are trivial on k. For, if∑a i D i = o, a i ∈ Rthen, since D i (x j )=δ i j , we geti∑o=( a i D i )x j = a j , j=1,...,n.iAlso, if D is any derivation of R which is trivial on k, then let Dx i =a i . Put ∑¯D= D− a i D i .Then∑¯Dx j = Dx j − ( a i D i )x j = owhich shows that since x 1 ,..., x n generate R, ¯D=o.ii

58 3. Algebraic function fieldsSuppose D is any derivation of k and let ¯D be an extension of D to R.Let ¯Dx i = a i . Let D o be another extension of D which has the propertyD o x i = a i , i=1,...,n.Then D−D o is a derivation of R which is trivial on k. But since(D−D o )x i = o, i=1,...,nit follows that D=D o . This gives us theTheorem 5. The derivations of K= k(x 1 ,..., x n ), the field of rationalfunctions of n variables over k which are trivial on k from a vector spaceof dimension n over K. A basis of this space of derivations is given bythe n partial derivationsD i = ∂∂x i, i=1,...,n.69Furthermore if D is any derivation of k, there exists only one extension¯D of D to K for which¯Dx i = a i , i=1,...,nwhere a 1 ,...,a n are any n quantities of K arbitrarily given.We now consider derivations of algebraic function fields.Let K = k(x 1 ,..., x m ) be a finitely generated extension of k. PutT = k[x 1 ,..., x m ]. In order to determine all the derivations of K, it isenough to determine the derivations of T since K is the quotient field ofT. Let D be a derivation of k; we wish to find extensions ¯D of D to K.Let R denote the ring of polynomials k[z 1 ,...,z m ] in m independentvariables. For any polynomial f (x 1 ,..., x m ) in T, denote by ∂ f the∂x ipolynomial obtained by substituting z i = x i , i=1,...,m in ∂ f¯where∂z if=f ¯ (z 1 ,...,z m ) is in R.If ∑f= a λ1 ,...,λ m x λ 11 ··· xλ m mλ

2. Derivations 59aλ 1 ,...,λ m ∈ k, put∑f D = (Da λ1 ,...,λ m )x λ 11 · xλ m m .λObviously f D is a polynomial in T. If ¯D is an extension of thederivation D, then clearly¯D f=f D +m∑i=1∂ f∂x i¯Dx ifor any f∈ T. Also ¯D is determined uniquely by its values on x 1 ,..., x m 70which generate T. Now ¯Dx i cannot be arbitrary elements of T. For, letY be the ideal in R of the set x 1 ,..., x m .Then for f (z 1 ,...,z m ) in Y ,f (x 1 ,..., x m )=o.Therefore, since ¯Do=0, the ¯Dx i would have to satisfy the infinityof equationsm∑0= f D ∂ f+ ¯Dx i∂x ifor every f in Y .Conversely, suppose u 1 ,...u m are m elements in T satisfyingf D +m∑i=1i=1∂ f∂x iu i = 0for every f in Y . For anyϕin T define ¯D by¯Dϕ=ϕ D + f D +m∑i=1∂ϕ∂x i¯Dx iwhere ¯Dx i = u i . Then clearly ¯D is a derivation of T and it coincideswith D on k. Furthermore ¯D does not depend on the wayϕis expressed

60 3. Algebraic function fieldsas a polynomial in x 1 ,..., x m . For, ifϕ=a(x 1 ,..., x m )=b(x 1 ,..., x m )then, since a−b∈Y have∑a D − b D (∂a+ u i − ∂b )u i = o∂x i ∂x iwhich proves our contention. Hencei71Theorem 6. Let K = k(x 1 ,..., x m ) and D a derivation of k. Let Ybe ideal in k[z 1 ,...,z m ] of the set x 1 ,..., x m . Let u 1 ,...,u m be anyelements of K. There exists a derivation ¯D and only one satisfying¯Dx i = u i , i=1,...,mand extending the derivation D in k, if and only if, for every f∈ Yf D +m∑i=1∂ f∂x iu i = o.and then for everyϕin K,¯Dϕ=ϕ D +m∑i=1∂ϕ∂x iu i .The infinite number of conditions above can be reduced to a finitenumber in the following manner. Since R=k[z 1 ,...,z m ] is a noetherianring, the ideal Y has a finite set f 1 ,..., f s of generators so that f∈ Ymay be writtens∑f= A i f i , A i ∈ Ri=1Suppose f 1 ,..., f s satisfy the above conditions, then since∑ ∑f D (x)= Ai D f i + fiD A i∂ f∂x i=i∑jA j∂ f j∂x i+i∑jf j∂A j∂x i,

2. Derivations 61we getf D (x)+m∑i=1∂ f∂x iu i = o.We may therefore replace the above by the finitely many conditionsf Di+m∑j=1∂ f∂x ju j = o, i=1,..., s.We now consider a few special cases. 72Let K= k(x) be a simple extension of k. Let D be a derivation of k.We will study extensions ¯D of D into K.(1) First let x be transcendental over k. The ideal of x in k[z] is zero.This means that we can prescribe ¯Dx arbitrarily. Thus for everyu∈Kthere exists one and only extension ¯D with¯Dx=u(2) Let now x be algebraic over k. Suppose x is inseparable overk. Let f (z) be the minimum polynomial of x in k[z]. Then f (z)generates the ideal of x in k[z]. But x being inseparable, f ′ (x)=o.This means that D has to satisfyf D = o.Also, ¯D is uniquely fixed as soon as we assign a value u to ¯Dx.This can be done arbitrarily as can be easily seen. Thus there existan infinity of extensions ¯D.(3) Finally, let X be separable. Then f (z), the irreducible polynomialof x over k is such thatf ′ (x)o.Since f (z) generates Y we must havef D (x)+ f ′ (x) ¯Dx=o,

62 3. Algebraic function fieldsor that ¯Dx is uniquely fixed by D.− ¯Dx= f D (x)f ′ (x)(*)73There is thus only one extension of D to K and it is given by (∗).In particular, K has no derivations, except the trivial one, over k.We shall now proveTheorem 7. In order that a finitely generated extension K = k(x 1 ,..., x n ) be separably algebraic over k, it is necessary and sufficient thatK have no non-trivial derivations over k.Proof. If K/k is algebraically separable, then since K is finitely generatedover k, it follows that K= k(x) for some x and the last of the considerationsabove shows that K has no nontrivial derivations over k. □74Suppose now K/k has no-trivial derivations. In case n = 1, ourconsiderations above show that K/k is separable. Let now n>1 andassume that theorem is proved for n−1 instead of n.PutK= K 1 (x n ) K 1 = k(x 1 ,..., x n−1 ).Then x n is separably algebraic over K 1 . For, if not, let x 1 be inseparableover K 1 or transcendental over K 1 . In both cases the zeroderivation in K 1 can be extended into a non-trivial deri-vation of K contradictionshypothesis over K.Thus x n is separable over K 1 . This implies, since K has no derivationsover k that K 1 and our theorem is proved.Note that in the theorem above, the fact that K/k is finitely generatedis essential. For instance, if k is an imperfect field and K= k p−∞ thenK/k is infinite. Also if a∈k then a=b p for some b∈K. If D is aderivation of K, thenDa=Db p = pb p−1 Db=oThis proves, in particular, that a perfect field of characteristic po,has only the trivial derivation.

2. Derivations 63If we take K to be the algebraic closure of the rational number filedthen K has only the trivial derivation.Let K= k(x, y) be an algebraic function field of one variable. Let usassume that x is a separating variable andϕ(X, Y) the irreducible polynomialof x, y over k. Then if D is a derivation of K over k,∂ϕ∂x Dx+∂ϕ ∂y Dy=oso that if we assume that y is separable over k(x), then ∂ϕ∂yDy= −∂ϕ ∂xDx∂ϕ∂yThis shows that the ratio Dy/Dx is independent of D.Also, for any rational functionψ(x, y) of x, ywhich gives,if DxoDψ= ∂ψ∂x Dx+∂ψ ∂y Dy( ∂ϕ )DψDx =∂ψ ∂x −∂ψ ∂x∂y ∂ϕ∂y o and hencewhich is a well known formula in elementary calculus.We shall now obtain a generalisation of theorem 7 to algebraic functionfields.Let K= k(x 1 ,..., x m ) be an algebraic function field of dimensionn, so that o≤n≤m. Let f 1 ,..., f s be a system of generators of the 75ideal Y of polynomials f (z 1 ,...,z m ) in k[x 1 ,..., x m ] which vanish for∂ fx 1 ,..., x m . Let for f in k[z 1 ,...,z m ] have the same meaning as∂x ibefore.Denote by M the matrixM=( ∂ f∂x i)i=1,...,m

64 3. Algebraic function fieldsj=1,..., swhere i is the row index and j the column index. We denote by t therank of the matrix M which is a matrix over K.Let V K (D) denote the vector space of derivations of K which aretrivial on k. This is a vector space over K. Denote by l the dimension ofV K (D) over K. We then haveTheorem 8.l+ t=m.Proof. For any integer p, denote by W p the vector space over K of dimensionp, generated by p-tuples (β 1 ,...,β p ),β i ∈ K. □Letσdenote the mappingσD=(Dx 1 ,..., Dx m )of V k (D) into W m . This is clearly a homomorphism of V k (D) into W m .The kernel of the homomorphism is the set of D for which Dx i = o; i=1,...,m. But since K is generated by x 1 ,..., x m , this implies that D=o.Thus V k (D) is isomorphic to the subspace of W m formed the vectors(Dx 1 ,..., Dx m ).76Consider now the vector space W s and let τ be the mapping(τ(α 1 ,...,α m )=(α 1 ,...,α m )M of W m into W s . Putβ j =m∑j=1α j∂ f i∂x j; i=1,..., sso that(β 1 ,...,β s )=(α 1 ,...,α m )M.The rank of the mappingτis clearly y equal to the rank t of thematrix M. It is the dimension of the image byτof W m into W s . Thekernel of the mappingτis the set of (α 1 ,...,α m )withβ i = o; i=1,..., s.

2. Derivations 65which, by theorem 6 is clearly isomorphic to the subspace of W m formedby vectors (Dx 1 ,..., Dx m ), D∈V K (D). This, by previous considerations,proves the theorem.We shall now prove theTheorem 9. With the same notations as before, there existlelements,say x 1 ,..., x 1 of dimension n over k such that K/k(x 1 ,..., x 1 ) is a separablyalgebraic extension.Proof. Since the matrix M has rank t, there exists a submatrix of M oft rows and which is non-singular. Choose notation in such a way, thatthis matrix is( ) ∂ f j i=m−t+1,...,mP= ,∂x i j= s−t+1,..., sNote that t≤ Min (s, m). Let L=k(x 1 ,..., x 1 ). ThenK=L(x 1+p ,..., x m ). Let D be a derivation of K over L. Then sincef j (x 1 ,..., x m )=o, we must have, by theorem 6m∑i=1∂ f j∂x iDx i = o.But since D is zero on L, 77ThusThis means thatDx i = o, i=1,...,l.m∑i=l+1⎛∂ f j∂x iDx i = o.Dx l+1P . ⎜⎝⎞⎟⎠ = .⎛⎜⎝ ⎟⎠Dx m oBut since|P|o, it follows that Dx l+1 = o...,Dx m = o whichshows that D=o⎞o

66 3. Algebraic function fieldsBut K=L(x l+1 ,..., x m ) is finitely generated over L. Using theorem7, it follows that K/L is algebraic and separable. We get incidentallyn≤1.We now prove the important□Theorem 10. Let K = k(x 1 ,..., x m ) be of dimension n. Then K isseparably generated over k, if and only if dim V K (D)=n. In that casethere exists, among x 1 ,..., x m , a separating base of n elements.Proof. If V K (D) has dimension n then theorem 9 shows that there exist nelements x 1 ,..., x n among x 1 ,..., x m such K/k(x 1 ,..., x n ) is separablyalgebraic.□78Suppose now that K/k is separably generated. Let y 1 ,...,y n bea separating base so that K/k(y 1 ,...,y n ) is separably algebraic. k(y 1 ,...,y n ) has n linearly independent derivations D 1 ,..., D n over k definedby⎧⎪⎨ o, if i jD i y j = ⎪⎩ 1, if i= jSince K/k(y 1 ,...,y n ) is separably algebraic and finite, it followsthat each of D 1 ,..., D n has a unique extension ¯D 1 ,..., ¯D n to K. Now¯D 1 ,..., ¯D n are linearly independent over K. For, if∑a i ¯D i = o, a i ∈ K,ithen ∑a i ¯D i (y j )=o for all j.iHence a j = o for j=1,...,n. Let now D be a derivation of K/k andlet Dy i = a i . Put ∑¯D=D− a i ¯D i .i

3. Rational function fields 67Then ¯Dy i = o for all i. Therefore since K/k(y 1 ,...,y n ) is separable,¯D=o.This proves thatdim V K (D)=n.and our theorem is completely established.Let K= k(x 1 ,..., x n , y) where k is of characteristic po and k isan imperfect field. Let y be algebraic over k and be a root ofz p − tt∈k. Then K/k is an inseparably generated extension andwhere n is the dimension of K/k.dim V K (D)=n+13 Rational function fieldsLet us now consider the field K = k(x) of rational functions of onevariable. Let y be any element of K. Hence y= f (x) where f and g areg(x)polynomials in x over k. Also K is the quotient field of the ring L=k[x]of polynomials in x.Assume that ( f (x), g(x)) = 1, that is that they have no factor incommon. Let n be defined by79n=max(deg f (x), deg g(x)).If n=o, then clearly f∈ k, g∈kand so y∈k. Let us assume thatno so that at least at least one of f and g is a non-constant polynomial.n is called the degree of y.Let F= k(y) be the field generated over k by y.Then x satisfies over F the polynomialϕ(z)= f (z)−yg(z).

68 3. Algebraic function fieldsϕ(z) is not a constant polynomial over k(y). For, letf (z)=g(z)=1∑a i z ii=oam∑ i , b i ∈ k.b i z⎫⎪⎬⎪⎭ii=oThen n=max(1, m). The coefficient of z n inϕ(z) is⎧a 1 if 1>m⎪⎨a 1 − yb 1 if 1=m⎪⎩ −yb m if 1

3. Rational function fields 69Let z in K be algebraic over k. Then z ∈ K = K 1 (x n ), K 1 =k(x 1 ,..., x n−1 ). Therefore by theorem 11 since z is algebraic over K 1 ,z∈ K 1 . By induction hypothesis, z∈k.ThusCorollary. If K= k(x 1 ,..., x n ) has dimension n over k, then k is algebraicallyclosed in KIt is easy to extend this to the case where K is a purely transcendentalextension of any transcendence degree.Since K= k(x), we call x a generator of K over k. Let y also be a 81generator so that K= k(y). Then(k(x) : k(y))=1which shows by our considerations leading to theorem 11 thaty= a(x)b(x)where a(x) and b(x) are coprime and have at most the degree 1 in x.Thusy= λx+µνx+ρwhereλ,µ,ν,ρare in k and since y is transcendental over k,λρ−µν0.An automorphism of K which is identity on k, is uniquely fixed byits effect on x. If it takes x into y then x and y are related as above. Ifx and y are related as above, then the mapping which assigns to x theelement y is an automorphism.If we consider the group of two rowed non-singular matrices, withelements in k, then each matrix gives rise to an automorphism of K/k.Obviously two matricesσandτgive rise to the same automorphism ifand only ifσ=λτ for someλo in k. HenceTheorem 12. The group of automorphisms of K= k(x) over k is isomorphicto the factor group of the group of two rowed matrices over kmodulo the group of matricesλE,λ0∈kand E is the unit matrix oforder 2.

70 3. Algebraic function fields82We shall call this group P 2 .From theorem 11, it follows that if L is an intermediary field betweenK and k then L is transcendental over k. But much more is true asin shown by the following theorem of Luroth.Theorem 13. If K= k(x) is a simple transcendental extension of k andk⊂L⊂K, L=K(ω) for someω∈K.Proof. We shall assume that Lk so that L is transcendental over kand contains an element t, transcendental over k andy by considerationsleading to theorem 11, we have K/k(t) is finite algebraic. Since L⊃k(t),it follows that(K : L)

3. Rational function fields 71so that since f (z) is irreducible, it follows that f (z) divides h(z)− wg(z),which is a polynomial of degree≤m in z. Let us therefore writeh(z)−wg(z)=f (x, z) ϕ(x, z)c 1 (x)b o (x) c o (x)whereϕ(x, z) is a primitive polynomial in z over k[x]. We therefore geton substituting the w= h(x)g(x) ,h(z)g(x)− g(z)h(x)= g(x)c 1(x). f (x, z)ϕ(x, z).b o (x)c o (x)The left hand side being a polynomial in x and z, f andϕbeingprimitive polynomials in z over k[x], it follows thath(z)g(x)− g(z)h(x)= f (x, z)ϕ 1 (x, z)whereϕ 1 (x, z)∈k[x, z]. We now compare degrees in x and z on bothsides of the above identity. On the right hand side the degree in x is≥ msince one of b o (x),...,b n (x) has degree m. Therefore the left side hasdegree in x≥m. But the degree in x equals degree of w≤m. Thusdegree of w=m. Since the left side is symmetrical in z and x,it followsthat it has degree m in z. Thereforeϕ 1 has to be independent of x.Hence h(z)−wg(z)= f (z)ϕ(z),ϕ(z) being independent of x. Thiscan happen only ifϕ(z) is a constant. This proves thatn=m.Now (K : k(w))=m=(K : L) and L⊃k(w). Thus 84L=k(w)and our theorem is proved.The analogue of Luroth’s theorem for K= k(x 1 ,..., x n ) is not knownfor n>1.Let K= k(x) and let G be a finite granite group of automorphism ofK/k. If L is the fixed field of G, then K/L is a finite extension of degreeequal to order of G. By Luroth’s theorem L=k(y) for some y. Thusdegree y= order of G.

72 3. Algebraic function fieldsFor instance, let G be the finite group of automorphisms of K= k(x)defined byx→ x, x→1− x, x→ 1 x , x→1− 1 x , x→ 11− x , x→ xx−1This is a group of order 6 and the fixed field will be k(y) where k(y)consists of all rational functions f (x) of x withf (x)= f (1− x)= f ( 1 x )= f (1− 1 x )= f ( 11− x )= f ( x1− x ).We have only to find a rational function of degree 6 which satisfiesthe above conditions. The functionf (x)= (x2 − x+1) 3x 2 (x−1) 2satisfies the above conditions and so y= f (x).Theorem 14. If G is any finite subgroup of P 2 of linear transformations85x→ ax+bcx+da, b, c, d∈k, ad− bc0 then there exists a rational function f (x) suchthat every functionϕ(x) which is invariant under G is a rational functionof f (x). f (x) is uniquely determined up to a linear transformationλ f (x)+µV f (x)+ρλ,µ,νρ∈k, λρ−µν0.We now consider the case of a rational function field K = k(x 1 ,... x n ) of n variables. Let G be a finite group of automorphisms of Kwhich are trivial on k and let L be the fixed field of G. Clearly L hastranscendence degree n over k. It is not known, except in simple cases,whether L is a purely transcendental extension of k or not. We shall,however, consider the the case where G is the symmetric group on n

3. Rational function fields 73symbols. So G≃S n . Let G operate on K in the following manner. Ifσis an element of S n , thenσis a permutation( )1, 2, 3,...,nσ= .σ 1 ,σ 2 ,σ 3 ,...,σ nWe defineσon K byσx i = x σi ; 1,...,n.We then obtain a faithful representation of S n on K and we denotethis group again by S n . An element of k which is fixed under S n andwhich therefore is in L is called a symmetric function of x 1 ,..., x n . Obviously(K : L)=n!and the galois group of K/L is S n . 86Consider the polynomialf (z)=(z− x 1 )...(z− x n ).Since every permutation in S n leaves f (z) fixed, it follows that f (z)∈L[z]. Let us writewheref (z)=z n − s 1 z n−1 + s 2 z n−2 −···+(1) n s ns i =∑1≤t 1

74 3. Algebraic function fieldsTheorem 15. Every rational symmetric function of x 1 ,..., x n is a rationalfunction over k of the elementary symmetric functions s 1 ,..., s n .Incidentally since L/k has dimension n, the elementary symmetricfunctions s 1 ,..., s n are algebraically independent over k.

Chapter 4Norm and Trace1 Norm and traceLet K/k be a finite extension and letω 1 ,...,ω n be a base of K/k so thateveryω∈Kmay be written∑ω= a i ω ia i ∈ k. By means of the regular representationi87ω→A ωwhere A ω = (a i j ) is an n− rowed square matrix with∑ωω i = a i j ω j i=1,...,njthe field K becomes isomorphic to the subalgebra formed by A ω in thealgebram n (k) of n rowed matrices over k. We denote by N K/k ω, S K/K ωthe norm and trace respectively ofω∈Kover k and they are defined byN K/k ω=|A ω |S K/k ω=trace A ω .75

76 4. Norm and TraceDefined as such, it follows thatN K/k ωω ′ = N K/k ω.N K/k ω ′S K/k (ω+ω ′ )=S K/k ω+S K/k ω ′forω,ω ′ ∈ K. Obviouslyω→N K/k ω is a homomorphism of K ∗ intoK ∗ and similarlyω→S K/k ω is a homomorphism of K + , the additivegroup, into k + .Letω ′ 1 ,...,ω′ n be any other basis of K/k. Then⎛⎜⎝ω ′ 1.ω ′ n⎞ω 1⎟⎠⎛⎜⎝= P .⎞⎟⎠ω n88where P is a non-singular matrix inm n (k). Since⎛ω 1⎛ω 1ω . ⎜⎝⎞⎟⎠ = A ω . ⎜⎝⎞⎟⎠ω n ω nit follows that⎛ω ′ 1ω ′ n⎞ω ′ 1ω . ⎜⎝ ⎟⎠ = PA′ ω⎛⎜⎝P−1 . ⎟⎠which shows that by means of the new basis the matrix associated toωis B ω whereB ω = PA ω P−1and then we haveω ′ n| B ω |=|A ω |Trace B ω = Trace A ω .This shows that N K/k ω and S K/k ω are invariantly defined and do notdepend on a basis of K/k.⎞

1. Norm and trace 77We writef K/k (x)=| xE− A ω |and call it the characteristic polynomial ofω. Obviously f K/k (0) =(−1) n | A ω | so thatwhereWe also see easily thatN K/k ω=(−1) n f K/k (0)=(−1) n a n . (1)S K/k ω=−a 1 (2)f K/k (x)= x n + a 1 x n−1 +···+a na 1 ,...,a n ∈ k.Let k⊂L⊂K be a tower of finite extensions. Let (K : L)=mand letΩ 1 ,...,Ω m be a basis of K/L. Similarly let (L : k)=n and let 89ω 1 ,...,ω n be a base of L/k. Then (ω 1 Ω 1 ,...,ω n Ω m ) is a base of K/k.Letω∈L and consider the matrix ofωby the regular representation ofK/k in terms of the base (ω 1 Ω 1 ,...,ω n Ω m ). Call it Ā ω .Then it is trivial to see that⎛A ω 0·········Ā ω =········· ⎜⎝⎞⎟⎠0 A ωa matrix of mn rows and columns. ThereforeN K/k ω=| Ā ω | (N K/k ω) (K:L)S K/k ω=(K : L)S K/k ωAlso the characteristic polynomials ofωas belonging to L and to Krespectively are f L/k (x) and f K/k (x) and they are related byf K/k (x)=( f L/k (x)) (K:L) , (3)In particular let L=k(ω). Then f L/k (x) is the minimum polynomialofω. We, therefore, have the

78 4. Norm and TraceTheorem 1. If K/k is a finite extension andω∈K,ϕ(x) its minimumpolynomial over k and f (x) its characteristic polynomial, thenwhere r=(K : k(ω)).f (x)=(ϕ(x)) r90From our formulae above, it follows that we can compute the normand trace ofωin K from a knowledge of its minimum polynomial.Let now K/k be a finite extension andωan element of K. Let[K : k(ω)] = m ′ , [k(ω) : k] = n ′ be the degrees of separability ofK over k(ω) and k(ω) over k respectively. Then K/k has m ′ n ′ distinctk-isomorphisms in an algebraic closureΩof k. Let { }σ i j ,i=1,...,n ′j=1,...,m ′ , bethese isomorphisms and let notation be so chosen thatσ i1 ,...,σ im , havethe same the same effect on k(ω). Then we may takeσ 11 ,σ 12 ,...,σ n ′ 1as a complete system of distinct isomorphisms of k(ω)/k inΩ.By our considerations above f k(ω)/k (x) is the polynomial ofωaswell as the characteristic polynomial ofωin k(ω)/k. If f K/k (x) is thecharacteristic polynomial ofωin k, thenf K/k (x)=( f k(ω)/k (x)) (K:k(ω)) (4)Now, because of the properties of the isomorphismsσ i j∏n ′ m ′i=1∏ ∏n ′(x−σ i j ω)= (x−σ i1 ω) m′ . (5)j=1But since f k(ω)/k (x) is the minimum polynomial ofω, we have⎧ ⎫n⎪⎨′ {k(ω):k}∏ ⎪⎬f k(ω)/k (x)= (x−σ ⎪⎩i1 ω) ⎪⎭i=1where{k(ω) : k}, as usual, denotes the degree of inseparability of k(ω)over k. Using (5) we get⎧ ⎫{k:k}⎧ ⎫⎪⎨∏n⎪⎬ ⎪⎨′ m∏′ {K:k}⎪⎬(x−σ i j (ω) = (x−σ⎪⎩ ⎪⎭⎪⎩i1 ω) . ⎪⎭i, ji=1i=1

1. Norm and trace 79But{K : k}={K : k(ω)}·{k(ω) : k} so that⎧ ⎫{K:k}⎪⎨∏ ⎪⎬(x−σ i j ω) = { f k(ω)/k (x) } m ′ {K:k(ω)}⎪⎩ ⎪⎭i, jwhich proves that 91{ ∏ {K:k}f K/k (x)= ((x−σω)}(6)σwhereσruns through all the distinct isomorphisms of k(ω) inΩ. Using(1) and (2) we get⎧ ⎫{K:k}⎪⎨∏ ⎪N K/k ω= ω ⎪⎩σ ⎬⎪ (7)⎭ω σ is a conjugate ofω. Similarly∑S K/k ω={K : k} ω σ . (8)σIf K/k is inseparable, then{K : k}= p t , t≥1 so that for everyω∈KS K/k ω=o.On the other hand, suppose K/k is finite and separable. Letσ 1 ,...,σ n be all the distinct isomorphisms of K/k inΩ, an algebraic closureof K. Then n=(K : k) and sinceσ 1 ,...,σ n are independent k-linearfunctions of K/k inΩ, it follows thatσσ=σ 1 +···+σ nis a non - trivial k−linear function of K/k inΩ. Therefore there exists aω∈Ksuch thatσω0. But by formula (8),so that we have theσω=ω σ 1+...+ω σ n= S K/k ω

80 4. Norm and TraceTheorem 2. A finite extension K/k is separable, if and only if there existin K an element whose trace over k in not zero.In case k has characteristic zero, or k has characteristic p∤n=(K : 92k), the unit element 1 in k has trace o. In order to obtain an elementωin K with S K/k ω0, in every case we proceed thus:Let K/k be separable and K= k(α) for an elementα. Letϕ(x) be itsirreducible polynomial over k andIt follows then thatϕ(x)=(x−α 1 )···(x−α n ).x n−1 ∑ϕ(x) =iα n−1iϕ ′ (α i )1x−α i.Comparing coefficients of x n−1 on both sides we get∑iα n−1iϕ ′ (α i ) = 1.If we putω= αn−1ϕ ′ (α) and observe thatϕ′ (α)∈K, we getS K/K ω=1.Using formula (6), it follows that if k⊂L⊂K is a tower of finiteextensions andω∈K, thenN K/k ω=N L/k (N K/L ω)S K/k ω=S L/k (S K/L ω)We now give a simple application of formula (7) to finite fields.Let k be a finite field of q= p a elements so that p is the characteristicof k. Let K be a finite extension of k so that (K : k)=n. Then K has

2. Discriminant 81q n elements. The galois group of K/k is cyclic of order n. Letσbe theFrobenius automorphism. Then forω∈K,ω σ =ω q .93The norm ofωisSince K ∗ has q n − 1 elements,σn−1N K/k ω=ω·ωσ·ωσ2···ωq n − 1=ω q−1 .α qn −1 = 1for allα∈K ∗ . Since K ∗ is a cyclic group, the number of elements in K ∗withq n − 1α q−1 = 1is precisely q n − 1/q−1, since q−1 divides q n − 1.Nowω→ N K/k ω is a homomorphism of K ∗ into k ∗ and the kernelof the homomorphism is the set ofωin K ∗ with1=N K/k ω=ωq n − 1q−1 .By the first homomorphism theorem we have, since k ∗ has only q−1elements, theTheorem 3. If K/k is finite and k is a finite field, then every non-zeroelement of k is the norm of exactly (K ∗ : k ∗ ) elements of K.It is clear that this theorem is not in general true if k is an infinitefield.

82 4. Norm and Trace2 DiscriminantLet K/k be a finite extension andω 1 ,...,ω n a basis of K/k. Supposeσ 94is a k-linear map of K into k, that isσ(ω)∈kσ(ω+ω ′ )=σω+σω ′σ(λ)=λσ(ω)whereω,ω ′ ∈ K,λ∈K. Let M σ denote the matrixM σ = (σ(ω i ω j))of n rows and columns. We denote by D K/k σ(ω 1 ,...,ω n ) its determinantand call it theσ- discriminant of the basisω 1 ,...,ω n of K/k. Ifω ′ 1 ,...,ω′ n is another basis, then⎛⎜⎝ω ′ 1.ω ′ n⎞ω 1⎟⎠⎛⎜⎝= P .⎞⎟⎠ω nwhere P is an n rowed non-singular matrix with elements in k.If P=(p i j ) then∑ω ′ i = p i j ω jjso that∑σ(ω ′ a ω′ b )= p ai p b j σ(ω i ω j )i, jwhich proves thatD σ K/k (ω′ 1 ,...,ω′ n)=|P| 2 D σ K/k (ω 1,...,ω n ).Therefore D σ K/k= 0 if it is zero for some basis.We now prove

2. Discriminant 8395Theorem 4. Ifω 1 ,...,ω n is a basis of K/k andσak-linear map of Kinto k, thenD σ K/k (ω 1,...,ω n )=0if and only ifσis the zero linear mapping.Proof. Ifσis the zero linear map, that is, one that assigns to everyelementωin K, the zero element, then D σ K/K = 0. Now let Dσ K/k = 0.This means that the matrix M σ with elements in k has determinant zero.Therefore there exist a 1 ,...,a n in k, not all zero, such that⎞a 1 0M⎛⎜⎝σ .⎞⎟⎠⎛⎜⎝= . ⎟⎠ .a n 0This means thatn∑σ(ω i ω j )a j = 0; 1=1,...,n.If we put z= ∑ jj=1a j ω j . then we haveσ(ω i z)=0, i=1,...,n.Letωbe any element in K. Since a 1 ,...,a n are not all zero, z0and so putωz = b 1ω 1 +···+b n ω n , b i ∈ k.( ω)Thenσ(ω)=σz· z = ∑ b i σ(ω i z)=o. This proves thatσis theitrivial map.The mappingω→S K/K ω is also a k-linear map of K into k. For abasisω 1 ,...,ω n of K/k we callD K/k (ω 1 ,...ω n )=|(S k/K (ω i ω j ))|the discriminant of the basisω 1 ,...ω n . Using theorem 2 and theorem 4,we get□

84 4. Norm and TraceTheorem 5. Discriminant of a base of K/k is not zero if and only if K/kis separable.96Let K/k be finite separable. Letσ 1 ,...,σ n be the distinct k-isomorphismsof K over k inΩ. Then∑S K/k ω= ω σ i.ThereforeD K/k (ω 1 ,...,ω n )=∣i1 , ..., ωσ n21.. (9)1 , ... ..., ωσ n∣1ω σ 11 , ωσ 2ω σ 1Since K/k is finite separable, K= k(ω) for someω. Also 1,ω,ω 2 ,...,ω n−1 form a base of K/k and we have1, ..., 1D(1,ω,...,ω n−1 )=ω σ 1, ..., ω σ 1..........................∣(ω n−1 ) σ 1, ..., (ω n−1 ) σ n∣Also, D(1,ω,...,ω n−1 ) / D(ω 1 ,...,ω n ) is the square of an elementof k. The determinant1, ..., 1ω σ 1, ..., ω σ n.........................∣(ω n−1 ) σ n, ..., (ω n−1 ) σ n∣is the called Van-der-Monde determinant. We call D(1,ω,...,ω n−1 ) isthe discriminant ofωand denote it by D K/k (ω).Let f (x) be the minimum polynomial ofω. Then f (x) = (x−ω σ 1)...(x−ω σ n). Since f ′ (ω)0 , we havef ′ (ω)=(ω σ 1−ω σ 2)(ω σ 1−ω σ 3).....(ω σ 1−ω σ n)and is an element of K. We call it the different ofωand denote it d K/k (ω).Also the Vander - monde determinant shows that

2. Discriminant 85D K/k (ω)=(−1) n(n−1)2 N K/k (d K/k ω)97Suppose K/k is a finite galois extension. Letσ 1 ,...,σ n be the distinctautomorphisms of K/k. We shall now proveTheorem 6. If k contains sufficiently many elements, then there existsin K an elementω such thatω σ 1,...,ω σ nform a basis of K over k.Proof. Itω∈Ksuch thatD(ω σ 1,...,ω σ n)othenω σ 1,...,ω σ nform a base of K/k. For,if∑a i ω σ i= o, a 1 ,...,a n ∈ kinot all zero, then sinceσ 1 ,...,σ n form a group∑a i ω σ jσ i= o, j=1,...,n.iTherefore from the expression (9) for D K/k (ω σ 1,...,ω σ n), it followsthat ⎞( S K/k (ω σ iω σ j)⎛⎜⎝) a 1 o.⎞⎟⎠⎛⎜⎝= . ⎟⎠a n oor that D(ω σ 1,...,ω σ n)=o which is a contradiction. We have thereforeto find anωwith this property. Putω σ i=x 1 ω σ i1 +···+ x nω σ in, i=1, 2,...where x 1 ,..., x n are indeterminates andω 1 ,...,ω n is a basis of K/k□ThenS K/k (ω σ iω σ j)=ω σ 1σ iω σ 1σ j,+···+ω σ nσ iω σ nσ j∑= S K/k (ω σ iaω σ jb )x ax b .a,b

86 4. Norm and TraceThen D K/k (ω σ 1,...,ω σ n) defined as the determinant of the matrix 98(S K/k (ω σ iω σ j)) is a polynomial in x 1 ,..., x n with coefficients in k.In order to prove that D K/k (ω σ 1,...,ω σ n) is non-zero polynomial,notice that by definition,⎛ω σ ⎞1ω. ⎜⎝ ⎟⎠⎛⎜⎝σ 1=1 , ωσ 12 , ..., ⎞⎛ωσ 1 xn 1......................ω σ n ω σ n1 , ωσ n2 , ..., ωσ n⎟⎠ . ⎜⎝⎞⎟⎠n x nso that ifω σ 1= 1 andω σ i= o for i>1, then x 1 ,..., x n are not allzero and for this set of values of x 1 ,..., x n , the polynomial D K/k (ω σ 1,...,ω σ n) has a valueo, as can be see from the fact that∣ ω σ 1, ..., ω σ n ∣∣∣∣∣∣∣∣ 2D K/k (ω σ 1,...,ω σ n)=....................∣ω σ nσ i, ..., ω σ nσ nTherefore if k has sufficiently many elements, there exist values in kof x 1 ,..., x n , not all zero, such that D K/k (ω σ 1,...,ω σ n)o.This proves the theoremIn particular, if k is an infinite field, there exists a base of K/k consistingof an element and its conjugates. Such a base is said to be anormal base.The theorem is also true if k is a finite field.

Chapter 5Composite extensions1 Kronecker product of Vector spacesLet V 1 and V 2 be two vector spaces over a field k and V 1 × V 2 theircartesian product. If W is any vector space over k, a bilinear functionf (x, y) on V 1 × V 2 is, by definition, a function on V 1 × V 2 into W suchthat for every x∈V 1 the mappingλ x : y→ f (x, y) is a linear functionon V 2 into W and for every y∈V 2 the functionµ y : x→ f (x, y) is alinear function on V 1 to W.A Vector space T over k is said to be a Kronecker product or tensorproduct of V 1 and V 2 over k, if there exists a bilinear functionθon V 1 ×V 2into T such that1) T is generated byθ(V 1 × V 2 )992) if V 3 is any vector space over k, then for every bilinear functionϕ onV 1 × V 2 into V 3 there exists a linear functionσ on T into V 3 such thatσθ=ϕ.This is shown by the commutative diagramWe shall now prove theϕV 1 × V 2 θ T87 σV 3

88 5. Composite extensionsTheorem 1. For any two vector spaces V 1 and V 2 over k there existsone and upto k-isomorphism only one Kronecker product T of V 1 andV 2 over k.100Proof. The uniqueness is easy to establish. For, let T 1 and T 2 be twovector spaces satisfying the conditions 1) and 2). Let T 1 be generatedbyθ 1 (V 1 × V 2 ) and T 2 byθ 2 (V 1 × V 2 ). Letσbe the linear map of T 1into T 2 such thatσ·θ 1 =θ 2 andτthe linear map of T 2 into T 1 such thatτ·θ 2 =θ 1 . Sinceθ 1 (V 1 × V 2 ) generates T 1 we see thatτ·σ is identityon T 1 . Similarlyσ·τ is identity on T 2 . Thusσandτare isomorphismsonto.□We now prove the existence of the space T.Let V be the vector space formed by finite linear combinations∑a xy (x, y)a xy ∈ k and (x, y)∈V 1 ×V 2 . Every bilinear function f on V 1 ×V 2 into V 3can be extended into a linear function f¯of V into V 3 by the prescriptionf ¯( ∑a xy (x, y) ) ∑= a x,y f (x, y)Let W be the subspace of V generated by elements of the type(x+ x 1 , y)−(x, y)−(x 1 , y)(x, y+y 1 )−(x, y)−(x, y 1 )(ax, y)−a(x, y)(x, by)−b(x, y)101where x, x 1 ∈ V 1 ; y, y 1 ∈ V 2 and a, b∈k. W is independent of V 3 .Also if f is a bilinear function on V 1 × V 2 , its extension f¯on V vanisheson W. Furthermore if f is any linear function on V vanishing on W,its restriction to V 1 × V 2 is a bilinear function. It is then clear that thespace of bilinear functions on V 1 ×V 2 is isomorphic to the space of linearfunctions on V which vanish on W. Ifσis any linear function on V/Winto V 3 andθthe natural homomorphism of V on V/W thenσ·θ is alinear function on V vanishing on W. Alsoσ→σ·θ is an isomorphism

1. Kronecker product of Vector spaces 89of the space of linear functions on V/W into V 3 on the space of linearfunctions on V vanishing on W.θ is clearly a bilinear function on V 1 ×V 2into T= V/W and furthermore, by definition of V, V/W is generated byθ(V 1 × V 2 ). Thus T is the required space.By taking k itself as a vector space over k, we haveTheorem 2. The space of bilinear functions on V 1 ×V 2 into k is isomorphicto the dual of the Kronecker product T.We denote by V 1 x○ k V 2 the kronecker product space. When there isno doubt about the field over which the kronecker product is taken wewill simply write V 1 x○V 2 .If T = V 1 x○V 2 we denote by x x○y the element in T which correspondsto (x, y) by the bilinear functionθon V 1 × V 2 into T. Sinceθ(V 1 × V 2 ) generates T, every element of T is of the form∑a x,y (x x○y) a xy ∈ k.Clearlyx x○o=o x○y=o x○o(x+ x⎧⎪⎨⎪⎩1 ) x○y= x x○y+ x 1 x○yx x○(y+y 1 )= x x○y+ x x○y 1ax x○y=a(x x○y)x x○ by = b(x x○y)with obvious notations. 102From our considerations it follows that in order to define a linearfunction on V 1 x○V 2 it is enough to define it on elements of the typex x○y. Also for every such linear function, there is a bilinear function onV 1 × V 2 .It is easy to see that the mapping x x○y→y x○x of V 1 x○V 2 to V 2 x○V 1is an isomorphism onto.Suppose now that V ∗ 1 and V∗ 2 are duals of V 1 and V 2 respectively overk. Ifσ∈V ∗ 1 andτ∈V∗ 2 and we define for (x, y)∈V 1× V 2 the functionσ.τ(x, y)=σx.τy,thenσ.τ is a bilinear function on V 1 × V 2 into k. We shall now prove the

90 5. Composite extensionsTheorem 3. If{x α } is a base of V 1 over k and{y β } a base of V 2 over k,then{x α x○y β } is a base of V 1 x○V 2 over k.Proof. We first prove that{x α x○y β } are linearly independent over k. Forall ∑ a αβ (x α x○y β )=ofor a αβ ∈ k, a αβ = o for all but a finite number ofα,β. By the method of constructing the tensor product, it follows that∑aαβ (x α , y β ) is an element of W. Letσandτbe elements of V ∗ 1 and V∗ 2defined respectively byσ(x α )=1ifα=γ= o otherwiseτ(y β )=1 if β=δ= o otherwise103□Thenσ·τ is a bilinear function on V 1 × V 2 and hence vanishes onW. Thusσ.τ (∑ a αβ (x α x○y β ) ) = o.But the left side equals a γδ . Thus all the coefficients a αβ vanish.Next any element of V 1 x○V 2 is a linear combination of elements ofthe type x x○y, x∈V 1 , y∈V 2 . But then x= ∑ a α x α and y= ∑ b β y β sothatx x○y= ( ∑ ) (∑ )a α x α x○ b β y βwhich equals ∑ a α b β (x α x○y β ). Our theorem is proved.We have incidentally theCorollary. If V 1 and V 2 are finite dimensional over k thendim V 1 x○V 2 = dim V 1· dim V 2 .Also since the dual of V ∗ 1 is isomorphic in a natural manner with V 1when dim·V 1 is finite, we get

1. Kronecker product of Vector spaces 91Corollary . If V 1 and V 2 are finite dimensional over k then V 1 x○ k V 2 isisomorphic to the dual of the space of bilinear functions on V 1 × V 2 intok.Let now A 1 and A 2 be two associative algebras over a field k. Wecan form the Kronecker product A=A 1 x○ k A 2 of the vector spaces A 1and A 2 over k. We shall now introduce a multiplication into A so as tomake it into an associative algebra.In order to do this, observe that the multiplication defined has to be 104a bilinear function on A× A into A. Since A is generated by elements ofthe type x x○y, it is enough to define this bilinear function on elements ofthe type (z, z 1 ) in A×Awhere z= x x○y and z 1 = x 1 x○y 1 . Putf (z, z 1 )=z·z 1 = x· x 1 x○y·y 1 .Now since (x, y)→ xx 1 x○yy 1 is a bilinear function on A 1 × A 2 intoA, by our previous considerations z→z·z 1 is a linear function on A intoA. Similarly z 1 → z·z 1 is a linear function on A. This proves that f isbilinear and that the multiplication so defined distributes addition. Thatthe multiplication is associative is trivial to see.A is called the Kronecker product algebra.We obtain some very simple consequences from the definition.α) If e 1 and e 2 are respectively the unit elements of the algebras A 1and A 2 then e 1 x○e 2 is the unit element of A 1 x○A 2 .For, since A=A 1 x○A 2 is generated by elements x x○y, it is enoughto verify (x x○y)(e 1 x○e 2 )=(e 1 x○e 2 )(x x○y). But this is trivial.β) If A 1 has x 1 ,..., x m as a base over k and A 2 has y 2 ,...,y n as abase over k then (x i x○y j ) is a base of A 1 x○A 2 over k. Furthermore if themultiplication tables for the bases are∑x i x j =ta (t)i j x t∑y i y j = b (t′ )i jy t ′t ′then ∑105(x p x○y q )(x r x○y s )= a (λ)pr b qs (µ) (x λ x○y µ )λ,µ

92 5. Composite extensionsγ) If A 1 and A 2 have unit elements e 1 and e 2 respectively then themappingsx→ x x○e 2y→e 1 x○yare isomorphisms of A 1 and A 2 into A. Thus A contains subalgebrasisomorphic to A 1 and A 2 .An important special case is the one where one of the algebras is afield. Let A be an algebra over k and K an extension field of k. Let A haveunit element e 1 and K the unit element e 2 . Form the Kronecker productA x○K over k. Then A x○K contains subalgebras A 1 and K 1 isomorphic toA and K respectively. For any x x○t in A x○K we havex x○t=x x○e 2 .e 1 x○t=e 1 x○t.x x○e 2so that A 1 and K 1 commute. Also every element of A x○K is of the form∑aαβ (x α x○t β ) where{x α } is a base of A over k and{t β } a base of K overk. But this expression can be written∑(∑a αβ (e 1 x○t β ) ) (x α x○e 2 ).αβ106This shows that A x○K is an algebra over K 1 with the base{x α x○e 2 }.If we identify A 1 with A and K 1 with K then A x○K can be considered asan algebra over K, a basis of A over k serving as a base of A x○K overK. A x○K is then called the algebra got from A by extending the groundfield k to K. We shall denote it by A K .It is clear that A K is commutative if and only if A is a commutativealgebra.Note. Even if A is a field over k, A K need not be a field over K.LetΓbe the rational number field andΓo=Γ( √ d) the quadraticfield overΓ. LetΓbe the real number field and consider the Kroneckerproduct A=Γ o x○¯Γ overΓ. Let e 1 , e 2 be a basis ofΓ o overΓwith themultiplication table,e 2 1 = e 1, e 1 e 2 = e 2 e 1 = e 2 , e 2 2 = de 1.

2. Composite fields 93The elements ofΓ o are of the form ae 1 +be 2 , a, b inΓ. The elementsof A are of the form ae 1 + be 2 with a, b∈ ¯Γ.Let first d>o. ThenΓ o x○¯Γ is not an integrity domain. For,( √ de 1 + e 2 )( √ de 1 − e 2 )=o.It can however be seen that A is then the direct sum of the two fieldsλ¯Γ andµ¯Γ whereλ= 1 2 (e 1+ e 2√d), µ= 1 2 (e 1− e 2√d).Let d

94 5. Composite extensionsObviously this is an equivalence relation and we can talk of a classof composite extensions.If k(K1 σ∪ Kτ 2 ) is a composite extension we denote by R(Kσ 1 ∪ Kτ 2 )the ring generated over k by K1 σ and Kτ 2inΩ. This ring is, in general,different from k(K1 σ∪ Kτ 2 ).Let now ¯K=K 1 x○K 2 be the Kronecker product of K 1 and K 2 . ¯Kcontains subfields isomorphic to K 1 and K 2 . We shall identify thesesubfields with K 1 and K 2 respectively. Let now k(K1 σ∪ Kτ 2) be a compositeextension and R(K1 σ∪ Kτ 2108 ) the ring of the composite extension.Define the mappingϕof ¯K into R(K1 σ∪ Kτ 2 ) byϕ ( ∑a xy (x x○y) ) =∑a xy x σ y τ ,where a xy ∈ k. Thenϕcoincides withσon K 1 and withτon K 2 .Sinceσandτare isomorphisms, it follows thatϕis a k-homomorphismof ¯K on R(K1 σ∪ Kτ 2). SinceΩis a field, it follows that kernel of thehomomorphism is a prime ideal Y of ¯K. Thus¯K/G≃ R(K σ 1 ∪ Kτ 2 ).If k(K1 σ′ ∪ Kτ′ 2) is another composite extension, thenµdefined earlier,is an isomorphism of R(K1 σ∪ Kτ 2) on R(Kσ′1∪ K2 τ′ ). Consider thehomomorphismϕ defined above. Define ¯ϕ on ¯K by ¯ϕ=µ·ϕ. We have∑ ( ) (∑¯ϕ( a xy x x○y) =µ a xy x σ y τ) ∑= a xy x σ′ y τ′ .Then ¯ϕ is a homomorphism of ¯K on R(K1 σ′ ∪ Kτ′ 2). But, sinceµis anisomorphism, it follows that ¯ϕ has G as the kernel. Thus the prime idealG is the same for a class of composite extensions.Conversely, if two composite extensions correspond to the sameprime ideal of ¯K, it can be seen that they are equivalent.We have, now, only to prove the existence of a composite extensionassociated with a prime ideal of ¯K. Let G be a prime ideal of ¯K andG ¯K. Since ¯K has a unit element, a prime ideal G ¯K always exists.Let A be the integrity domainA= ¯K/G

2. Composite fields 95109Letϕbe the natural homomorphism of ¯K on A. Then K ϕ 1 and Kϕ 2are subfields of A. Since ¯K is generated by elements of the type x x○y,K ϕ 1 and Kϕ 2 , are different from zero. Clearly A=R(Kϕ 1 ∪ Kϕ 2). Hencethe quotient field of A is a composite extension. Hence for every primeideal G ¯K, there exists a composite extension. We have hence provedtheTheorem 4. The classes of composite extensions of K 1 and K 2 standin (1,1) correspondence with the prime ideal G ¯K of the Kroneckerproduct ¯K of K 1 and K 2 over k.K τ 2Consider now the special case where K 2 /k is algebraic. Let k(K1 σ∪) be a composite extension. Thenk(K σ 1 ∪ Kτ 2 )⊃R(Kσ 1 ∪ Kτ 2 )⊃Kσ 1 .Since every element of K 2 is algebraic over k, k(K1 σ∪Kτ 2) is algebraicover K1 σ. This means that R(Kσ 1 ∪ Kτ 2) is a field and so coincides withk(K1 σ∪ Kτ 2). ThusTheorem 5. If K/k is algebraic, then every prime ideal G ¯K of K isa maximal ideal.Let K/k be an algebraic extension and L/k any extension. The Kroneckerproduct ¯K= K x○ k L is the extended algebra (K) L of K by extendingk to L. ¯K is thus an algebra (commutative) over L. If G is a primeof ¯K, then by above, it is a maximal ideal and ¯K/G gives a compositeextension. Since ¯K is an algebra over L we may regard ¯K/G as anextension field of L.Let now G 1 ,...,G m be m distinct maximal ideals of ¯K, none of them 110equal to ¯K. Let L i = ¯K/G i be a composite extension. Form the directsum algebra ∑as a commutative algebra over L. We shall now proveiL i

96 5. Composite extensionsTheorem 6.∑i¯K/G i ≃ ¯K/ ⋂ iG iProof. We shall construct a homomorphismϕ of ¯K on ∑ i L i and showthat the kernel is ⋂ G i . □iIn order to do this let us denote byσ i the natural homomorphism of¯K on L i (this is identity on L), i=1,...,m. If x∈ ¯K, thenσ i x∈L i .Defineϕon ¯K by ∑ϕ(x)= σ i x.That this is a homomorphism on ¯K is easily seen; for, if x, y∈ ¯K∑ ∑ ∑ϕ(x+y)= σ i (x+y)= σ i x+ σ i y=ϕx+ϕy.i∑ ∑ ∑ ∑ϕ(xy)= σ i (xy)= (σ i x)(σ i y)=( σ i x)( σ i y)=ϕxϕy.iiiiiii111The kernel of the homomorphism is set of x such thatϕx=o. Thusσ i x=oso that x∈G i for all i. Hence x∈∩ i G i . But every y∈∩ i G i hasthe propertyϕy=o. Thus the kernel is precisely ⋂ G i .iWe have only to prove that the homomorphism is onto.To this end, notice that for each i, i=1,...,m, there is a b i ∈ ¯K with⎧⎪ ⎨∈ G j , jib i⎪ ⎩ G iFor, since G i and G j , ji are distinct, there is a j ∈ G j which isnot in G i . Put b i = ∏ a j . Then b i satisfies above conditions since G i isi jmaximal.Let now ∑ c i be an element in the direct sum, c i ∈ L i . By definitioniof b i⎧⎪ ⎨= o if jiσ j b i⎪ ⎩ o if j=i

3. Applications 97Since L i is a field, there exists x i o in L i such thatx i σ i b i = o i .σ i being a homomorphism of ¯K on L i , let y i ∈ ¯K withσ i y i = x i . Put∑c= b i y iThen∑∑∑ϕ(c)= σ j (b i y i )=ijiwhich proves the theorem completely.Suppose in particular K/k is finite. Then K L has over L the degree(K : k). Since, for a maximal ideal, G ¯K, ¯K/G has over L at most thedegree (K : k), we getic i1≤(¯K/G i : L)≤(K : k)i=1,...,m.This means that ¯K has only finitely many maximal ideals and∑¯K/ ⋂ G≃ ¯K/GGGthe summations running through all maximal ideals of ¯K. 112Thus K and L have over k only finitely many inequivalent compositeextensions.3 ApplicationsThroughout this sectionΩ will denote an algebraically closed extensionof k and K and L will be two intermediary fields betweenΩ and k. Acomposite of K and L inΩwill be the field generated over k by K andL. It will be denoted by KL.Let ¯K be the Kronecker product over k of K and L. There is, then, ahomomorphismϕof ¯K on R(K∪ L) given byϕ ( ∑a xy (x x○y) ) ∑= a xy x·y.

98 5. Composite extensionsSuppose that this homomorphism is an isomorphism of ¯K onto R(K∪L). This means that∑ ∑a xy (x x○y)=o⇐⇒ a xy x·y=oor that every set of elements of K which are linearly independent over kare also so over L. Incidentally, this givesK∩ L=k.Conversely, suppose K and L have the property that every set ofelements of K which are linearly independent over k are also so overL. Then the mappingϕis an isomorphism of ¯K on R(K∪L). For, if∑axy xy=owe express x and y in terms of a base of L/k and a base of113K/k giving∑b αβ x α y β = oBut this means all b αβ are zero. Thereforeϕis an isomorphism.It shows that every set of elements of L which are linearly independentover k are also so over K.We call two such fields L and K linearly disjoint over k. Note thatL∩ K= k. We deduce immediately1) If L and K are are linearly disjoint over k then any intermediaryfield of K/k and any intermediary field of L/k are also linearly disjoint.Suppose now that K/k is algebraic. Then every prime ideals of ¯K ismaximal. Let, in addition, K and L be linearly disjoint over k. Since ¯Kis isomorphic to R(K∪ L), it follows that (o) is a maximal ideal. Hence¯K is a field. Thus2) If K/k is algebraic and K and L are linearly disjoint over k, thereexists but one class of composite extensions of K and L over k.Let K/k be a finite extension. Then, for some maximal ideal G , ¯K/Gis isomorphic to KL. Since ¯K/G may be considered as an extension fieldof L, we get ( ¯K/G : L)≤(K : k), that is(KL : L)≤(K : k)Clearly if G = (o), equality exists, and then K and L are linearlydisjoint over k. The converse is true, by above considerations. Hence,in particular,

3. Applications 993) If (K : k)=m and (L : k)=n, then(KL : k)≤mn;equality occurs if and only if K and L are linearly disjoint over k. 114We now consider the important case, K/k galois. By the considerationsabove, it follows that KL/k is algebraic over L. SinceΩis aalgebraically closed, it contains the algebraic closure of KL. Letσbean isomorphism of KL inΩ, which is identity on L. Its restriction to Kis an isomorphism of K inΩ. ButΩcontains the algebraic closure ofK and henceσK=K. Since KL is generated by K and L, it followsthatσKL⊂KL. Hence KL/L is a galois extension. (KL/L is alreadyseparable since elements of K are separable over k).We now prove theΩKL K L K∩ LkTheorem 7. If K/k is a galois extension and L, any extension of k thenG(KL/L)≃G(K/K∩ L)Proof. In the first place, we shall prove that K and L are linearly disjointover K∩ L. For this, it is therefore enough to prove that every finite sety 1 ,...,y m of elements of L which are linearly independent over K∩ L,are also so over K.□x iIf possible, let y 1 ,...,y m be dependent over K so that ∑ x i y i = o,∈ K. Let us assume that y 1 ,...,y m are dependent over K but no

100 5. Composite extensionsproper subset of them is linearly dependent over K. Therefore, all the x iare different from zero.We may assume x 1 = 1. Letσbe an element of G(KL/L). Then 115∑ ∑0=σ( x i y i )= σx i·σy iBy subtraction we get, sinceσ1=1,∑0= (x i −σx i )y iii1This means that x i =σx i , i=2,...,m. But, sinceσis arbitrary inG(KL/L), it follows that x i ∈ L. But x i ∈ K. Thus y 1 ,...,y m are linearlydependent on K∩L which is a contradiction. Hence K and L are linearlydisjoint over K∩ L.Therefore, KL is isomorphic to the Kronecker product of K and Lover K∩ L.Supposeσis any L-automorphism of KL. Its restriction to K isan automorphism of K and leaves K∩ L fixed. Consider the mappingσ→ ¯σ of G(KL/L) into G(K/K∩ L). This is clearly a homomorphism.If ¯σ is identity element of G(K/K∩L), thenσis identity on K. Since it isalready identity on L, it is identity on KL. Thus, G(KL/L) is isomorphicto a subgroup of G(K/K∩ L). To see that this isomorphism is ontoG(K/K∩ L), letτεG(K/K∩ L). Any element of KL may be written(since K and L are linearly disjoint) in the from,∑x α y ααwhere x α are linearly independent elements of K over K∩ L and y α ∈ L.This expression is unique. Extendτto ¯τ in G(KL/L) by defining⎛∑ ∑¯τ ⎜⎝ x α y α⎞⎟⎠ = y α τ(x α ).ααi116This is well defined; for, if ∑ α y α τ(x α ) = 0, then, since{x α } arelinearly independent over K∩ L,{τ(x α )} are also linearly independent

3. Applications 101over K∩ L and since K and L are linearly disjoint over K∩ L, all y α arezero. Thus ¯τ is an automorphism of KL/L.Our theorem is thus proved.In particular, if K and L are both galois extensions of k and K∩L=kthen, by above,G(KL/L)≃G(K/k)G(KL/K)≃G(L/k)Also, since KL/k is algebraic, letσbe an isomorphism (trivial onk) of KL inΩ. Since its restrictions on K and L are auto morphisms, itfollows that KL/k is a galois extension. We now proveTheorem 8. G(KL/k) is isomorphic to the direct product of G(K/k) andG(L/k).Proof. G(K/k) and G(KL/L) are isomorphic and for every elementσ inG(K/k), by the previous theorem, we have the extension ¯σ, an elementof G(KL/L) determined uniquely byσ. Similarly, ifτ ∈ G(L/k), ¯τdenotes its unique extension into an element of G(KL/k). KKLLk□We now consider the mapping(σ,τ)→ ¯σ¯τof the direct product G(K/k)·G(L/k) into G(KL/k).Supposeλis an element of G(KL/k). Its restrictionλto K is an 117element of G(K/k). Considerλ¯1 . Now ¯λ −1 λ is identity on K. By the1

102 5. Composite extensionsisomorphism of G(KL/k) and G(L/k), this defines a unique elementµ 1of G(L/k). Henceλ= ¯λ 1 ¯µ 1 .Thus the mapping above is a mapping of the direct product G(K/k)·G(L/k) onto G(KL/k). We have only to prove that is a homomorphismto obtain the theorem. It is clearly seen thatλis identity if and only ifλ 1 andµ 1 are identity.Letσ,σ ′ be two elements of G(K/k) andτ,τ ′ in G(L/k). Letλandµ be the unique elements in G(KL/k) defined byλ= ¯σ¯τ, µ= ¯σ ′ , ¯τ ′ .Consider to elementλµ. Its restriction to K isσσ ′ and its restrictionto L isττ ′ . Thusλµ=σσ ′·ττ ′which proves that the mapping is a homomorphism.The theorem is now completely proved.

Chapter 6Special algebraic extensions1 Roots of unity118Consider the polynomial x m − 1 in k[k], where k is a field. Let k havecharacteristic p. If p=o, then x m − 1 is a separable polynomial overk, whereas if po, the derivative of x m − 1 is mx m−1 which is zero,if p divides m. If p∤m, then x m − 1 is a separable polynomial overk. Therefore, let m>o be an arbitrary positive integer, if k has characteristiczero and m, an integer prime to p, if k has characteristic p0.Then x m − 1 is a separable polynomial over k and it has m roots inΩ, analgebraic closure of k. We call these m roots, the mth roots of unity.Ifρandτare mth roots of unity thenρ m = 1 = τ m . Therefore(ρτ) m =ρ m τ m = 1, (ρ −1 ) m = (ρ m ) −1 = 1 which shows that the mth rootsof unity from a group. This group G m is abelian and of order m.Let now d/m. Any root of x d − 1 inΩis also a root of x m − 1.Ifρisan mth root of unity such thatρ d = 1, thenρis a root of x d − 1. Sincex d − 1 has exactly d roots inΩ, it follows that G m has the property thatfor every divisor d of m, the order of G m , there exist exactly d elementsof G m whose orders divide d. Such a group G m is clearly a cyclic group.HenceThe mth roots of unity form a cyclic group G m of order m. 119There exists, therefore, a generatorρ of G m .ρ is called a primitivemth root of unity. Clearly, the number of primitive mth roots of unity is103

104 6. Special algebraic extensionsϕ(m). All the primitive mth roots of unity are given byρ a with 1≤a

2. Cyclotomic extensions 105Any x in R may be written as a µ nwhere a is an integer. Define themappingσas followsσx=ρ a n ,so thatσis a function on R with values in H. The mapping is welldefined: for if x= bµ m, thenµ m a=µ n b. Suppose m>n; thenb=aν n+1···ν mso thatρ b m =ρaν n+1···ν mm =ρ a n by choice ofρ n. We now verify thatσis ahomomorphism of R on H. If x a , y= b are in R and m≥n, thenµ n µ mσ(x+y)=σ( aν n+1···ν m + b)=ρ aν n+1···ν m +bmµ mwhich equalsρ a n·ρ b m=σx·σy. Also, since any root of unity is in someH n , it is of the formρ a n so that, for x= a µ n,σx=ρ a n. We have, therefore,to determine the kernel of the homomorphism. It is the set of x in R suchthatσx=1.If x= a µ n, then 1=σx=ρ a n so that|µ n |a and so x is an integer. The 121converse being trivial, it follows that the kernel is precisely the additivegroup of integers and our theorem is established.2 Cyclotomic extensionsLet k be a field and x m − 1 a separable polynomial in k[x]. This implies,in case k has characteristic p o, that p does not divide m. Letρbe a primitive mth root of unity inΩ, an algebraic closure of k. ThenK = k(ρ) is the splitting field of x m − 1 inΩ. Therefore, K/k is aseparable, normal extension. Let G be its galois group. Ifσ∈G,σisdetermined by its effect onρ. Sinceρis a primitive mth root of unity, soisσρ. For,(σρ) m =σ(ρ m )=1;

106 6. Special algebraic extensionssoσρ is a root of unity. Also, if (σρ) t = 1, thenσ(ρ t )=1. Sinceσisan automorphism, it follows thatρ t = 1 or m/t. Thusσρ=ρ ν , (ν, m)=1.Ifσ,τare in G, letσρ=ρ ν , (ν, m)=1 andτρ=ρ µ (µ, m)=1.Thenστ(ρ)=σ(τρ)=σ(ρ µ )=ρ µνwhich shows thatστ=τσ or that G is abelian.Consider now the mappingg :σ→ν122whereσρ=ρ ν , (ν, m)=1. This is clearly a homomorphism of G intothe multiplicative group prime residue classes mod m. The kernel ofthe mapping g is set ofσwithσρ=ρ.But thenσt=t for all t ∈ K, so that by, galois theory,σ is theidentity.Let us call extension K= k(ρ) a cyclotomic extension of k. We thenhave proved theTheorem 2. The Cyclotomic extension k(ρ)/k is an abelian extensionwhose galois is isomorphic to a subgroup of the group of prime residueclasses mod m whereρ m = 1 andρis a primitive mth root of unity.LetΓbe the prime field contained in k. ThenΓ(ρ) is a subfield ofk(ρ)=K. Let G be the galois group K/k.K= k(ρ)kΓ(ρ)ΓLetσbe in G and ¯σ the restriction ofσtoΓ(ρ). Sinceσis identityon k andσρ is again a primitive root of unity, it follows that ¯σ is an

2. Cyclotomic extensions 107automorphism ofΓ(ρ)/Γ. It is easy to see that the mappingσ→ ¯σ is anisomorphism of G into the galois group ofΓ(ρ)/Γ.We shall therefore confine ourselves to studying the galois group ofΓ(ρ)/Γ.First, letΓbe the rational number field andρaprimitive mth root ofunity. LetΓ(ρ) be the cyclotomic extension. Let f (x) be the primitiveintegral polynomial which is irreducible inΓ[x] an whichρsatisfies.Then f (x) is a monic polynomial. For, since f (x) divides x m−1 ,x m − 1= f (x)ψ(x).ψ(x) has rational coefficients and soψ(x)= a b ψ 1(x) whereψ 1 (x) is a 123primitive integral polynomial and a and b are integers. From the thetheorem of Gauss on primitive polynomials it follows that f (x) is monic.Let p be a prime not dividing m. Letϕ(x) be the minimum polynomial(which is monic and integral) ofρ p . We assert that f (x)=ϕ(x).For, if not, f (x) andϕ(x) are coprime and sox m − 1= f (x)·ϕ(x)·h(x).for some monic integral polynomials h(x).Consider the polynomialϕ(x p ). It hasϕas a root and so f (x) dividesϕ(x p ). Henceϕ(x p )= f (x)g(x),g(x), again, a monic and integral polynomial. Considering the abovemod p we getf (x)g(x)≡ϕ(x p )≡(ϕ(x)) p (mod p)so that f (x) divides (ϕ(x)) p ( mod p). If t(x) is a common factor off (x) and (ϕ(p)( mod p) then (t(x)) 2 divides x m − 1 mod p, which isimpossible, since p∤m and x m − 1 does not have x as a factor. Thus ourassumption f (x)ϕ(x) is false.This means that, for every prime p ∤ m,ρ p is a root of f (x). If(ν, m)=1, thenν= p 1 p 2··· p 1 where p 1 ,..., p 1 are primes not dividing

108 6. Special algebraic extensions124m. By using the above fact successively, we see that, for everyν, (ν, m)=1,ρ ν is a root of f (x). Therefore∏ρ m (x)= (x−ρ ν ) (1)(ν,m)=1divides f (x). Butϕ m (x) is fixed under all automorphisms ofΓ(ρ)/ Γ sothat f (x)=ϕ m (x). We have proved theTheorem 3. IfΓis the field of rational numbers andρis a primitive mthroot of unity, then the galois group of the cyclotomic extensionΓ(ρ)/Γis isomorphic to the group of prime residue classes mod m. The irreduciblepolynomialϕ m (x) ofρis given by (1).ϕ m (x) is called the cyclotomic polynomial of order m. Its degree isϕ(m). In order to be able to obtain an expression forϕ m (x) in terms ofpolynomials overΓ, we proceed thus.We introduce the Mobius function defined as follows:It is a functionµ(n) defined for all positive integers n such that1) µ(1)=12) µ(p 1··· p t )=(−1) t where p 1 ,..., p t are distinct primes.3) µ(m)=oif p 2 /m, p being a prime.From this, one deduces easily4) µ(m)·µ(n)=µ(mn) for (m, n)=1.For, one has to verify it only for m= p 1··· p t , n=q 1···q 1 wherep ′ s and q ′ s are all distinct primes. Thenµ(m)=(−1) t ,µ(n)=(−1) 1andµ(mn)=(−1) t+1 .We now prove the following simple formula5)∑µ(d)=oif m>1d|m= 1 if m=1125the summation running through all divisors d of m.

2. Cyclotomic extensions 109If m = 1, the formula reduces to (1). So, let m > 1. Let m =p a 11 ··· pa tt be the prime factor decomposition of m. Any divisor d of mis of the form p b 11 ··· pb tt where o≤b i ≤ a i , i=1,...,t. In view of (3),it is enough to consider divisors d of m for which o≤b i ≤ 1. In thatcase,∑ t∑µ(d)= ( t i )(−1)i = o.d|mi=oLet now f (n) be a function defined on positive integers, with valuesin a multiplicative abelian group. Let g(n) also be such a function. Wethen have the Möbius inversion formula,∏f (d)=g(n)⇐⇒ f (n)= ∏ d|n(g(d)) µ( n d )d|nSuppose ∏ f (d)=g(n). Thend|n∏ ∏(∏µ(d).(g(d)) µ( n g ) = f (d 1 ))d|nd|nd i | n dChanging the order products, we get∏( f (d1 ) )∑ µ(d);d 1 |n d| d n1using formula (5), we obtain the inversion formula. The converse followsin the same way.Consider now the integers mod m. Divide them into classes in thefollowing manner. Two integers a, b are in the same class if and only if(a, m)=(b, m).Let d/m and C d , the class of integers a ( mod m) with (a, m)=d. 126Then a is of the form dλ where (λ, m d )=1. Thus C d hasϕ( m d ) elements.The classes C d , for d|m, exhaust the set of integersprimitive mth root of unity, thenm∏x m − 1= (x−ρ t ).t=1mod m. Ifρis a

110 6. Special algebraic extensionsIn view of the above remarks, we can write∏(∏)x m − 1= (x−ρ t ).t∈C dd|mBut if t∈C d ,ρ t =ρ dλ , (λ, m d )=1 and soρt is a primitive m dof unity. Using the definition ofϕ m (x), if follows that∏x m − 1= ϕ d (x)d|mth rootBy the inversion formula we getϕ m (x)= ∏ d|m(x d − 1) µ( m d )Comparison of degrees on both sides gives the formula∑ϕ(m)= dµ( m ∑d )=md|md|mµ(d)d .We may computeϕ m (x) for a few special values of m. Let m= p, aprime number. Thenϕ p (x) xp − 1x−1 = xp − 1+···+ x+1.If m= pq, the product of two distinct primes, thenϕ pq (x)= (xpq − 1)(x−1)(x p − 1)(x q − 1) .127Let us now consider the case whenΓis the prime filed of p elements.Obviously,Γ(ρ)/Γ is a cyclic extension. If we define the cyclotomicpolynomial as before, it is no longer irreducible overΓ[x]. For instance,let p=5and m=12. Thenϕ 12 (x)= (x12 − 1)(x 2 − 1)(x 6 − 1)(x 4 − 1) = x4 − x 2 + 1.

2. Cyclotomic extensions 111Alsox 4 − x 2 + 1=(x 2 − 2x−1)(x 2 + 2x−1) (mod 5).Therefore,Γ(ρ)/Γ has degree

112 6. Special algebraic extensionsTheorem 5. A division ring with a finite number of elements is a filed.Proof. Let D be the division ring and k its centre. Then k is a field. Dbeing finite, let k have q elements. If D is of rank n over k, then D hasq n elements. We shall prove that n=1. □Let D 1 be a subalgebra of D over k. Let D 1 have rank m over k.Then D 1 has q m elements. But D ∗ 1 is a sub group of D∗ so that q m − 1divides q n − 1. This means that m|n. For, if n=tm+µ, o≤µ(q−1) ϕ(n)dx

3. Cohomology 1133 CohomologyLet G be a finite group and A an abelian group on which G acts as agroup of left operators. Let A be a multiplicative group. We denote elementsof G byσ,τ,ρ,..., and elements of A by a, b, c,.... We denotesby a σ the effect ofσon a. Then(ab) σ = a σ b σ(a τ ) σ = a στ 130Let 1 be the unit element of A. Denote by G n , n≥1 the Cartesianproduct of G with itself n times.A function on G n with values in A is said to be an n dimensionalCochain or, simply, an n cochain. This function f (x 1 ,..., x n ) has valuesin A. We denote by aσ 1 ,...,σ n the element in A which is the valuetaken by the n cochain for valuesσ 1 ,...,σ n of its variables. We denotethe function also, by aσ 1 ,...,σ n . If aσ 1 ,...,σ n and bσ 1 ,...,σ n aretwo functions, we define their product bySimilarlyo σ1 ,...,σ n= a σ1 ,...,σ n· b σ1 ,...,σ n.o σ1 ,...,σ n= (a σ1 ,...,σ n) −1is called the inverse of a σ1 ,...,σ n. With these definitions, the n cochainsform a group C n (G, A) or simply C n .We define C o (G, A), the zero dimensional cochains, to be the groupof functions with constant values, that is functions a σ on G such thata σ = ais the same for allσ∈G. It is then clear that C o is isomorphic to A.We now introduce a coboundary operator∂ in the following manner.∂(=∂ n ) is a homomorphism of C n into C n+1 defined bya σ1 ,...,σ n+1=∂a σ1 ,...,σ n

114 6. Special algebraic extensions= a σ 1σ 2 ,...,σ n+1n∏(a σ1 ,...,σ i−1 ,σ i σ i+1 ,...,σ n+1) (−1)i (a σ1 ,...,σ n) (−1)n+1 .n=1For n ≥ o, we define the groups Z n (G, A) and B n+1 (G, A) inthe following manner. Z n (G, A) is the kernel of the homomorphismC n ∂n−→ C n+1 and B n+1 (G, A) is the homomorphic image. The elementsof Z n (G, A) are called n dimensional cocycles or n-cocycles and theelements of B n+1 (G, A) are n+1-dimensional coboundaries or n+1coboundaries. The homomorphism∂has the property131∂∂= identity (2)which proves that B n+1 (G, A) is a subgroup of Z n+1 (G, A) and we canform for every n>o the factor group H n (G, A) the n-dimensional cohomologygroup. We verify (2) only in case n=o and 1 which are theones of use in our work.The coboundary of a zero cochain is a one cochain given byIts coboundary, by definition is∂a=a σ = aσ a . (3)132∂∂a=a σ,τ = a σ· a σ τa στ.Substituting from (3), we get a σ,τ = 1 which verifies (2) for n=o.We define the zero coboundary, that is the elements in B o (G, A), tobe the function with value 1 on allσ∈G. Thus B o (G, A) consists onlyof the identity. An element in Z o (G, A) will be the constant function awitha σ = afor allσ. HenceH o (G, A) is isomorphic with the set of a∈A with the property a σ =a for allσ∈G.A one dimensional cocycle is a function a σ for which∂a σ = 1. But∂a σ = a σ· a σ τa στ.

3. Cohomology 115Therefore, the elements in Z ′ (G, A) are functions a σ witha σ a σ τ = a στ.A one coboundary is, already, a function a σ of the form a σ /a. It is,certainly, a one cocycle.For our purposes, we shall need also the additive cohomology, insteadof the multiplicative cohomology above. We regard now A + asan additive, instead of a multiplicative, group. Then C n (G, A + ) is anadditive abelian group. We define the coboundary operator asa σ1 ,...,σ n+1=∂a σ1 ,...,σ n= a σ 1σ 2 ,...,σ n+1+n∑(−1) i a σi ,...,σ i−1 ,σ i ,τσ i+1 ,...,σ n+1+ (−1) n+1 a σi ,...,σ n.i=1As before, a zero cochain is a constant function on G and its coboundarya σ is∂a=a σ = a σ − a.A one cochain a σ is a cocycle ifwhich is the same thing as∂a σ = 0a σ + a σ τ = a στ.Exactly as before, we see that H o (G, A), the zero dimensional additivecohomology group is isomorphic to the set of elements a in A + with 133a σ = a for allσ.We now consider the case when G is a cyclic group. Letσbe a generatorof G so that 1,σ,σ 2 ,...,σ n−1 are all the elements of G. Takingmultiplicative cohomology, if a σ is a one cocycle, thena τµ = a τ a τ µor a τ µ= a τµ /a τ . Substituting forτsuccessively 1,σ,...,σ n−1 we geta 1+σ+···+σn−1µ = 1.

116 6. Special algebraic extensionsWe denote a 1+σ+···+σn−1µ by Na µ and call it the norm of a µ . Thus, ifa µ is a cocycle, then,Na µ = 1.Conversely, suppose a is an element in A withNa=a 1+σ+..+σn−1 =1 = 1.We can define a cocycle a µ such that a σ = a. For letµ=σ ν , forsomeν. Puta µ = a 1+σ+···+σν−1 .Obviously, a σ = a. Also, a σ is a cocycle. For, ifτ=σ ν′ .a µ a µ τ= a 1+σ+···+σν−1 (a 1+σ+···+σν′ −1) σν= a 1+σ+···+σν−1 +σ ν +···+σ ν+ν′ −1= a µτ .134In a similar manner, for additive cohomology, we have a τµ = a τ +a τ µwhere a µ is a cocycle. If G is cyclic, on substituting 1,σ,...,σ n−1 forτ, we getS a µ = a µ + a σ µ+···+a σn−1µ = o.We call S a µ the spur or trace of a µ . If a is any element of A, withtrace S a=a+a σ +···+a σn−1 = o, then the cocyclewhereµ=σ ν , has the propertya µ = a+a σ +···+a σν−1a σ = a.We now apply the considerations above, in the following situation.K/k is a finite galois extension with galois group G. Then G acts onK ∗ and also the additive group K + as a group of operators. We might,therefore, consider the cohomology groups H o (G, K ∗ ), H 1 (G, K ∗ ),...and H o (G, K + ), H 1 (G, K + )... etc. As before, H o (G, K ∗ ) and H o (G, K + )are isomorphic to subgroups of K ∗ and K + with a σ = a for allσ. This,by galois theory, shows

3. Cohomology 117Theorem 6. H o (G, K ∗ ) is isomorphic to k ∗ and H o (G, K + ) is isomorphicto k + .But what we are interested in, is the following important theoremdue to Artin.Theorem 7. The group H 1 (G, K ∗ ) and H 1 (G, K + ) are trivial.Proof. Let us first consider multiplicative cohomology. If a σ is a cocycle,we have to prove that it is a coboundary. The elementsσ,τ,... ofG are independent k-linear mappings of K intoΩ, the algebraic closure.Hence, if (a σ ) are elements of K ∗ ,∑a σ·σσis a non-trivial k-linear map of K intoΩ. Therefore, there exists aθo 135in K such that ∑a σ θ σ o.σ□Put b −1 = ∑ a σ θ σ = ∑σ τa τ θ τ . ThenTherefore∑(b −1 ) σ = a σ τ θστ .a σ∑b σ= a σ a σ τθ στ .Since (a σ ) is a cocycle, we geta σ∑ ∑b σ= a στ θ στ = a τ θ τ = b −1 .ττττThusa σ = bσ b = bσ−1which is a coboundary.

118 6. Special algebraic extensions136Consider now the additive cohomology. K/k being finite and separable,there exists aθ∈Ksuch that∑θ σ = S K/k θ=1.Put nowσ∑−b= a σ θ σa σ being an additive cocycle. Then∑−b σ = a σ τθ στ .But a σ = a σ· 1= ∑ τ a σ·θ τ so that∑ ∑a σ − b σ = (a σ + a σ τ )θ στ = a στ θ στ =−bτστwhich proves that a σ = b σ − b is a coboundary.Our theorem is completely proved.We apply the theorem in the special case where G is cyclic. Letσbe a generator of G. If a σ is a cocycle then, in multiplicative theoryτa 1+σ+···+σn−1σ = 1or N K/k a σ = 1. Similarly in additive cohomology,a σ + a σ σ +···+aσn−1 σ = oor S K/k a σ = o.Using theorem 7, we obtain ’theorem 90’ of Hilbert.Theorem 8. If K/k is a finite cyclic extension,σ, a generator of thegalois group of K/k and a and b two elements of K with N K/k a=1 andS K/k b=0 respectively, thenfor two elements c, d in K.a=c 1−σb=d σ − d

4. Cyclic extensions 1194 Cyclic extensionsLet K/k be a cyclic extension of degree m. Put m=np a , (n, p)=1 if pis the characteristic of k; otherwise, let m=n.Let G be the galois group of K/k. It has only one sub-group of orderp a . Let L be its fixed field. Then K/L is cyclic of degree p a and L/k iscyclic of degree n prime to p. Letρbe a primitive nth root of unity andk(ρ) the cyclotomic extension. The composite F=Lk(ρ) is cyclic overk(ρ) and of degree prime to p. We shall see that K over L and F overk(ρ) can be described in a simple manner.We shall, therefore, consider the following case, first.K/k is a cyclic extension of degree m and p∤m, if k has character- 137istic po; otherwise, m is an arbitrary integer. Also, k contains all themth roots of unity. We then have the theorem of Lagrange.Theorem 9. K= k(w) where w m ∈ k.Proof. Letρbe a primitive mth root of unity.ρ is in k.□Hence, since K/k has degree m,N K/k ρ=ρ m = 1.By Hilbert’s theorem, therefore, ifσis a generator of the galoisgroup of K/k, then there is anω∈Ksuch thatω 1−σ =ρ.Sinceρis a primitive mth root of unity,ω,ω σ ,ω σ2 ,... are all distinctand are conjugates ofω. Hence our theorem.ω satisfies a polynomial x m − a, a∈k. If K= k(ω ′ ), whereω ′ alsosatisfies a polynomial x m − b, b∈k,thenω ′σ =ω ′·ρ + ,whereρ t is a primitive mth root of unity and so (t, m)=1.Now ( ω′) σω t = ω′·ρtω t·ρt = ω′ω t

120 6. Special algebraic extensionswhich shows thatω ′ =ω t· c, c∈k.We shall call x m − a a normed polynomial. We have, then, the138Corollary. If k(ω)=K= k(ω ′ ), whereωandω ′ are roots of normedpolynomials, thenω ′ =ω t· c,where (t, m)=1 and c∈k.We consider the special case, m=q, a prime number. Let K/k be acyclic extension of degree q. Let K= k(α) and letα 1 = (α),α 2 ,...,α qbe the irreducible polynomial ofαover k. Supposeσis a generator ofthe galois group of K/k.Let notation be so chosen thatα σ 1 =α 2,α σ 2 =α 3,...,α σ q−1 =α q,α σ q=α 1 .Since k contains the qth roots of unity and every qth root of unityρ1 is primitive, we construct the Lagrange Resolvent.Thenω=ω(α,ρ)=α 1 +ρα 2 +···+ρ q−1 α q .ω σ =α 2 +ρα 3 +···+ρ q−2 α q +ρ q−1 α 1which shows thatω σ =ρ −1 ω. Hence K= k(ω). Also,(ω q ) σ =ω qwhich proves thatω q ∈ k andωsatisfies a normed polynomial.In particular, if k has characteristic2, and K/k has degree 2, thenK= k( √ d)for d∈k.The polynomial x q − a for a∈k is, thus, either irreducible and, then,a root of it generates a cyclic extension, or else, x q − a is a product oflinear factors in k.

4. Cyclic extensions 121139We study the corresponding situation when K has characteristic po and K/k is a cyclic extension of degree p t , t≥1.We first consider cyclic extensions of degree p.Let K/k be a cyclic extension of degree p andσ, a generator of thegalois group of K/k. Letµbe a generic element of the prime fieldΓcontained in k.Introduce the operator P x= x p − x. ThenP(x+µ)=P x.Also, the onlyαin k satisfying Pα=o are the elements ofΓandthese are all the roots of P x=o.The element 1 in k has the propertyS K/k 1=o,so that by Hilbert’s theorem, there is anω∈Ksuch that1=ω σ −ω.Therefore, since K/k has degree p,K= k(ω).In order to determine the polynomial of whichωis a root, considerPω.(Pω) σ = Pω σ = P(ω+1)=Pωwhich shows that Pω∈k. If we put Pω=α, thenωis a root theirreducible polynomialx p − x−αin k[x]. ω is a root of the polynomial andω σ =ω+1. Sinceσis agenerator of the galois group of K/k, the roots of x p − x−α areω,ω+1,...,ω+ p−1.A polynomial of the type x p − x−α,α ∈ k is called a normedpolynomial.Suppose K=k(ω ′ ) whereω ′ also satisfies a normed polynomial. 140

122 6. Special algebraic extensionsThenω ′ ,ω ′ + 1,...,ω ′ + p− 1 are the roots of this normed polynomial.Soω ′σ =ω ′ + h, o

4. Cyclic extensions 123elements ofΓserve the same purpose as the roots of unity in the firstcase.If k has characteristic 2 and K/k is a separable extension of degree2, thenK= k(ω)whereω 2 −ω∈k.Observe, also, that any polynomial x p − x−α, forα∈k, is eitherirreducible over k and so generates a cyclic extension over k, or splitscompletely into linear factors in k. For, ifωis a root of x p − x−α then,forµ∈Γ,ω+µ is also a root.Just as we denote a root of x m −α by m√ α, we shall denote, forα∈k,by α P , a root of xp − x−α. It is obvious thatone value of α , all the values arePαPω,ω+1,...,ω+ p−1.is p valued and ifωisWe now study the case of cyclic extension of degree p n , n≥1.Let K/k be a cyclic extension of degree p n andσ, a generator ofthe galois group G of K/k. Since G is cyclic of order p n , it has onlyone subgroup of order p and hence, there exists a unique subfield L of Ksuch that K/L is cyclic of degree p and L/k is cyclic of degree p n−1 = m.Let us assume that n≥2.σ m is of order p and hence is a generator of the galois group of K/L. 142Thus K= L(ω) whereσ m ω=ω+1andωsatisfies Pω=α∈L.Putσω−ω=β. Thenσ m β=σ(σ m ω)−σ m ω=β which shows thatβ∈L. Also,S L/k β=β+β σ +···+β σm−1 .Substituting the value ofβ, we getS L/k β=(σω−ω)+(σ 2 ω−σω)+···+(σ m ω−σ m−1 ω)=σ m ω−ω=1.

124 6. Special algebraic extensionsHenceβhas the propertyS L/k β=1.It is easy to see thatα is not k. For,Pβ=σ(Pω)−Pω=σα−αandαin k would mean Pβ=o orβ∈Γ. This means that S L/k β=o.We now proceed in the opposite direction. Let L be cyclic of degreep n−1 > 1 over k. We shall construct an extension K which is cyclic overk and contains L as a proper subfield. Letσbe a generator of the galoisgroup of L/k.Let us chooseβ∈L, such thatS L/k β=1.Now S L/k (Pβ)=P(S L/k β)=o which shows thatPβ=σα−α143for someαin L. Also,αis not in k. We claim that forλ∈k,x p − x−α−λ (4)is irreducible over L[x]. For, if it is not irreducible, it is completelyreducible in L. Letωbe a root of x p − x−α−λ in L. ThenThis means thatThusω p −ω=α+λ.P(σω−ω)=σ(Pω)−Pω=σα−α=Pβ.P(σω−ω−β)=oorσω−ω−β=µ. Since S L/k (σω−ω)=o and S L/k µ=o, it follows thatS L/k β=o, which is a contradiction. Thus forλ∈k, (4) is irreducible

4. Cyclic extensions 125in L[x]. Letωbe a root of this irreducible polynomial for someλ. ThenK= L(ω)/L is cyclic of degree p.Let ¯σ denote an extension ofσto an isomorphism of K/k inΩ, thealgebraic closure of k. Then ¯σ is not identity on L. Since Pω=α+λandλ∈k, we getNow¯σ(Pω)=σα+λ.P( ¯σω−ω)= ¯σ(Pω)−Pω=σα−α=Pβand, again, we haveP( ¯σω−ω−β)=oor that ¯σω=ω+β+µ which shows that ¯σ is an automorphism of K/k.If t is any integer,¯σ t ω=ω+β+β σ +···+β σt−1 + tµ.This shows that 1, ¯σ, ¯σ 2 ,..., ¯σ pn−1 have all different effects onω,so that they are distinct automorphisms of K/k. But, since K/k hasdegree p n , it follows that K/k is cyclic of degree p n . We have, hence,the important144Theorem 11. If k is a field of characteristic p and admits a cyclic extensionK of k containing L and of degree p m , m>n, for every m.In fact, if k is an infinite field, we may veryλover k and obtainan infinity of extensions K over k with the said property. It followstheorem 10.Corollary. If k is a field of characteristic p and admits a cyclic extensionof degree p n for some n ≥ 1, then its algebraic closure is of infinitedegree over it.

126 6. Special algebraic extensions5 Artin-Schreier theorem145We had obtained, in the previous section, a sufficient condition on a fieldso that its algebraic closure may be of infinite degree over it. We wouldlike to know if there are fields K which are such that their algebraicclosures are finite over them. The complete answer to this question isgiven by the following beautiful theorem due to Artin and Schreier.Theorem 12. IfΩit an algebraically closed field and K is a subfieldsuch that1

5. Artin-Schreier theorem 127whereω q 1 =ω∈L.4) Any irreducible polynomial over L is either linear or of degree q. 146For, if t is its degree andα, a root of it then t=(L(α) : L) dividesq. Thus, every polynomial in L[x] splits in L into product of linearfactors and of factors of degree q.Consider, in particular, the polynomial x q2 −ω. InΩ, we can writex q2 −ω=π µ (x−µ q2√ ω) (5)whereµruns through all q 2 th roots of unity. Since (Ω : L)>1, thepolynomial x q2 −ω has, in L[x], an irreducible factor of degree q.Since this factor is formed by q of the linear factors on the right of(5), this factor is of the formx q +···+ε ε√ ω,ε being a q 2 th root of unity. We assert that this is a primitive q 2 throot of unity. For if not, it is either 1 or a primitive qth root of unity.Sinceε q√ ω∈L, it would then follow thatq√ ω∈L. Therefore we getΩ=L(ε).5) LetΓbe the prime field contained in L. Consider the fieldΓ(ε).LetΓ(ε) contain the q ν th but not q ν+1 roots of unity. This integercertainly exists. For, if L has characteristic zeroν=2. If L hascharacteristic p (and this is different from q, from (2)), thenΓ(ε)contains p f elements where f is the smallest positive integer withp f ≡ 1( mod q 2 ). If q ν is the largest power of q dividing p f − 1 thenΓ(ε) contains the q ν th, but not the q ν+1 the roots of unity. 147Letρbe a primitive q v+1 th root of unity. Thenρsatisfies a polynomialof degree q(irreducible) over L. ThusΓ(ρ) is of degree q overΓ(ρ)∩ L. Now,ρbeing a primitive q ν+1 th root of unityρ q is a q ν throot of unity and so is contained inΓ(ε). But all the roots of x q −ρ qare q v+1 th roots of unity. Thus x q −ρ q is the irreducible polynomial

128 6. Special algebraic extensionsofρoverΓ(ε). HenceΓ(ρ)Γ(ε)Γ(ρ)∩ LΓand (Γ(ρ) :Γ(ε))=q=(Γ(ρ) : (Γ(ρ)∩L).IfΓis a finite field, then (Γ(ρ)/Γ is cyclic and it cannot have twodistinct subfieldsΓ(ε) andΓ(ρ)∩L over which it has the same degree.SinceεL, it follows thatΓ(ε) andΓ(ρ)∩L are distinct and so,Γhas characteristic zero. In this case, if q is odd,Γ(ρ)/Γ is cyclic andthe same thing holds. Thus q=2. We know, then thatν=2. If idenotes a fourth root of unity, thenΩ=L(i).6) Form above, it follows that n=2 t for some t and K does not containthe 4th root of unity. Now t=1, for if not, let t>1. Let M be asubfield of L such that (L : M)=2. Then (Ω : M(i))=2 and, byabove considerations, M(i) cannot contain 1 which is a contradiction.SoK(i)=Ω.148We shall make use of this theorem in the next chapter.□6 Kummer extensionsWe now study the structure of finite abelian extensions of a field k. Weuse the following notation:-k is a field of characteristic p, not necessarily>o.n is a positive integer not divisible by p if po, otherwise arbitrary.α, the group of non-zero elements of k. A generic element ofαwillalso be denoted, following Hasse and Witt, byα.

6. Kummer extensions 129α n is the group of nth powers of elements ofα.ω, a subgroup ofαcontainingα n and such that(ω :α n )

130 6. Special algebraic extensions3) Λ /α ≃ω /α n. (These are isomorphic groups).Consider the homomorphismΛ→Λ n ofΛinto itself. It takesαintoα n . The kernels inΛandαare both the same. SinceΛis taken toω by this homomorphism, it follows thatΛ/α≃ω/α n . Incidentally,therefore,Λ/α is a finite group.We shall now prove the important property,4) G is isomorphic toω/α n .(This proves that K/k is a finite abelian extension.)In order to prove this, consider the ’pairing’, (τ,Λ) of G andΛ, givenby(σ,Λ)=Λ 1−σ , (7)150σ∈G,Λ∈Λ. Because of definition ofΛ,(σ,Λ) n = Λn(Λ n ) σ= 1.Thus (σ,Λ) is an nth root of unity. Also,(στ,Λ)= Λ Λ ( ) σ ΛΛ στ= Λ σ= Λ τ = Λ Λ σ· ΛΛτ= (σ,Λ)·(τ,Λ).Furthermore,(σ,ΛΛ ′ )= Λ′ Λ ′(Λ ′ Λ) σ= Λ Λ σ,Λ′ Λ σ= (σ,Λ)(σ,Λ′ ).Thus (σ,Λ) defined by (7) is a pairing.Let G o be the subgroup of G consisting of allσwith (σ,Λ)=1, forallΛ. SinceΛis generated byΛ 1 ,Λ 2 ,...,Λ t ,α we getΛ i =Λ σ i .ButΛ 1 ,...,Λ t generate K. Therefore,β=β σ for allβ∈K.By galois theory,σ=1. Thus G o = (1).LetΛ o be the subgroup of allΛwith (σ,Λ)=1, for allσ. Forthe same reason as before,Λ o =α. Thus, by theorem on pairing, G isisomorphic toΛ/ α . (4) is thus proved.

6. Kummer extensions 131Observe, now, that sinceΛ n ⊂α, it follows the G is an abelian groupwhose exponent divides n.We will denote K symbolically by K= k( τ√ ω).We shall now proveTheorem 13. Let K/k be a finite extension with abelian galois group Gof exponent dividing n, then K= k( n√ ω) for a sub groupω⊂α withω /α n finite.Proof. Let ¯Λ denote the group of non-zero elements of K with the prop- 151ertyΛ n ∈ α forΛ ∈ ¯Λ. Then, by considering the homomorphismΛ→Λ n , it follows thatΛ ¯ /α ≃ω /α n,whereω= ¯Λ n . We shall prove that K= k( n√ ¯ω). In order to prove this,it suffices to prove thatG≃ω /α n. (8)For, in that case, construct the field k( n√ ω). Because of the propertiesof K, it follows that K⊃ k( n√ ¯ω). But, by the previous results, (k( n√ ω) :k)=order ofω /α n=order of G.HenceK= k( n√ ω).We shall, therefore, prove (8).Let G∗ denote the dual of G. For everyΛin ¯Λ, define the functionχ Λ (σ)=Λ 1−σon G into ¯Λ. SinceΛ n ∈α, it follows thatχ Λ (σ) is an nth root of unity.Also,χ Λ (στ)= Λ Λ ( ) σ ΛΛ στ= Λ σ Λ τ = Λ Λ σ· ΛΛ τ=χ Λ(σ)·χ Λ (τ)so thatχ Λ is a character of G.□

132 6. Special algebraic extensionsLetχbe any character of G. Since the exponent of G divides n,χ n (σ)=1 for allσ. Therefore,χ(σ) is an nth root of unity and hence isan element of k. Also, sinceχis a character of G,χ(στ)=χ(σ)χ(τ)152which equalsχ(σ)·(χ(τ)) σ . Thusχ(σ) is a one cocycle and, by theorem7,χ(σ)=β 1−σforβ∈K. But (χ(σ)) n = 1 so thatβ n = (β n ) σfor allσorβ n ∈α. This means, by definition of ¯Λ thatβ∈ ¯Λ.Consider the mappingω→χ Λ of ¯Λ into G ∗ . This is, trivially ahomomorphism. By above, it is a homomorphism onto. The kernel ofthe homomorphism is set ofΛfor whichχ Λ (σ)=1 for allσ. That isΛ=Λ σfor allσ. By galois theoryΛ∈α. But every element inαsatisfies thiscondition. Thusαis precisely the kernel, and¯Λ /α ≃ G ∗ .Since G is a finite abelian group, G≃ G ∗ and our theorem is completelyproved.We call a finite abelian extension K/k, a Kummer extension if1) k contains the nth roots of unity, p∤n, if p( o) is characteristic ofk,2) K/k has a galois Galois group of exponent dividing n.What we havethen proved isTheorem 14. The Kummer extensions of k stand in (1, 1) correspondencewith subgroupsω ofαwithω /α n finite.

7. Abelian extensions of exponent p 1331537 Abelian extensions of exponent pWe had introduced earlier the operator P which is defined byP x= x p − x.Let k be a field of characteristic p and denote by k + the additivegroup of k. Thenα→Pα is a homomorphism of k + into itself. Thekernel of the homomorphism is precisely the set of elements inΛ, theprime field of characteristic p, contained in k.Ifα∈k, we denote by α P , a root of the polynomial xp − x−α.Obviously α Pis p valued. Also,α+βP = α P + β P . (9)Let us denote by Pk + , the subgroup of k + formed by elements Pα,α∈k + . Letωbe a subgroup of k + with the properties,k + ⊃ω⊃Pk +ω/ Pk + is finite.Let K be the extension field of k, formed by adjoining to k all theelements α , froα∈ω. We denotePK= k( ω P ).We then obtain in exactly the same way as before1) K/k is a finite abelian extension.2) The galois group G of K/k is isomorphic withω/ Pk +.3) G is an abelian group of exponent p. For the proofs, we have to useadditive, instead of multiplicative, pairing and the property (9).

134 6. Special algebraic extensionsSuppose, now, on the other hand, K/k is a finite abelian extension of 154exponent p, where p is the characteristic of k.ThenK= k( ω P ),whereωis a subgroup of k + withω/Pk + finite. For the proof of this,we have to use additive instead of multiplicative, cohomology. For, letG be the galois group of K/k and G ∗ its character group. Denote, by¯Λ, the additive subgroup of K + formed by elementsΛwith PΛ∈k + .Denote, byω, the group P ¯Λ. ThenDefineχ Λ (σ) byThen¯Λ /k +≃ω /Pk +χ Λ (σ)=Λ−Λ σ .P(χ Λ (σ))=PΛ−(PΛ) σ ).But, PΛ∈k + by definition. Hence, P(χ Λ (σ))=o.Therefore,χ Λ (σ) is an element ofΓ. The rest of the proof goesthrough in the same way and we have theTheorem 15. If k has characteristic po, the finite abelian extensionsK/k of exponent p stand in a(1, 1) correspondence with subgroupsωofk + such thatω/Pk + is finite, and then K= k( ω P ).8 Solvable extensions155We propose to study now the main problem of the theory of algebraicequations, namely the ‘solution’ of algebraic equations by radicalsk will denote a field of characteristic p,(p=o or po),Ωwill be itsalgebraic closure and n, n 1 , n 2 ,... integers> o which are prime to p, if khas characteristic po, otherwise arbitrary. An elementω∈Ω will besaid to be a simple radical over k ifω n ∈ k, for some integer n. k(ω)/kis then said to be a simple radical extension. k(ω)/k is clearly separable.Ifωis a root of x p − x−a, for a∈ k,ω is called a simple pseudo radical

8. Solvable extensions 135over k.k(ω)/k is a pseudo radical extension and is separable. In fact, itis a cyclic extension over k. This situation occurs, only if po.An extension field K/k is said to be a generalized radical extensionif it is a finite towerk=K o ⊂ K 1 ⊂ K 2 ⊂···⊂K m = Kwhere K i /K i−1 is either a simple radical or a simple pseudo radical extension.Every element it K is called a generalized radical. A typicalelement would be√n√a11 + n 2a 2 + a 3P +..Clearly, K/k is a separable extension. A generalized radical extensionis called a radical extension if K i /K i−1 is a simple radical extensionfor every i. A typical element of K would, then ben 1√a 1 + n 2 √ a 2 +···.Let f (x) be a polynomial in k[x] and let it be separable. Let K beits splitting field. Then K/k is a galois extension. The galois groupG of K/k is called the group of the polynomial f (x). The polynomialf (x) is said to be solvable by generalized radicals, if K is a subfield 156of a generalized radical extension. It is, then, clear that the roots off (x) are generalized radicals. In order to prove the main theorem aboutsolvability of a polynomial by generalized radicals, we first prove somelemmas.Lemma 1. Every generalized radical extension is a subfield of a generalizedradical extension which is galois, with a solvable galois group.Proof. Let K/k be a generalized radical extension so thatk=K o ⊂ K 1 ⊂···⊂ K m = K,where K i /K i−1 is a simple radical or simple pseudo radical extension.Let K i = K i−1 (ω i ). Then eitherω n i∈ K i−1 for some n i or Pω i ∈ K i−1 .Let (K : k)=n. Put N= n, if k has characteristic zero; otherwise, letn=N·p a ,

136 6. Special algebraic extensionsa≥o, (p, N)=1. Letρbe a primitive Nth root of unity.□Let L 1 = K o (ρ). The L 1 /K o is a simple radical extension. SinceK 1 = K o (ω 1 ), put L 2 = K o (ρ,ω 1 ). Then L 2 is the splitting field of thepolynomial (x N −1)(x n1 −a 1 ) or (x N −1)(P x−a 1 ) depending on whetherω 1 is a simple radical or a simple pseudo radical. In any case, L 2 /K o isa galois extension. FurthermoreL 2 = L 1 (ω 1 )157so that L 2 /L 1 is cyclic, since, whenω 1 is a simple radical, L 1 containsthe requisite roots of unity. Letσ 1 ,...,σ l be the distinct automorphismsof L 2 /K o . Putl∏f (x)= (x n2 − a σi2 ),i=1ifω 2 is a simple radical withω n 22 = a 2, andf (x)=l∏(P x−a σ i2 ),i=1ifω 2 is a simple pseudo radical with Pω 2 = a 2 .Then f (x) is a polynomial in k[x]. Let L 3 be its splitting field. ThenL 3 is galois over k. Also, L 3 is splitting field of f (x) over L 3 so thatL 3 /L 2 is either a Kummer extension or else, an abelian extension ofexponent p. In this way, one constructs a galois extension T of k suchthatL o = k⊂L 1 ⊂ L 2 ⊂ L 3 ⊂···⊂L m−1 ⊂ T= L m ,where L i /L i−1 is either a kummer extension or an abelian extension ofexponent p. Clearly, L i /L i−1 and, hence, T/k is a generalized radicalextension. Let G i , i=o, 1, 2,...,m be the galois group of T/L i . ThenG= G o ⊃ G 1 ⊃......⊃G m = (e)is a normal series. Further, by our construction, G i−1 /G i is the galoisgroup of L i /L i−1 and hence, abelian. Thus, G is a solvable group. Thelemma is thus proved.

8. Solvable extensions 137Lemma 2. If K/k is a finite solvable extension, then K is a subfield of ageneralized extension over k.Proof. Let G be the galois group of K/k and G solvable. Let n be theorder of G and putn= p a Nwith the same connotation, as before. Letρbe a primitive Nth root of 158unity and L=k(ρ). Then L/k is a simple radical extension. Let M bea composite of K and L. Then M/L is a galois extension with a galoisgroup which is isomorphic to a subgroup of G and hence, solvable. LetG o be the galois group of M/L and let it have a composition seriesG o ⊃ G 1 ⊃···⊃G m = (e).Then G i /G i+1 is a cyclic group of prime degree. Let L o = L,L 1 ,...L m = M be the fixed fields of G o , G 1 ,..., G m respectively. ThenL i /L i−1 is a cyclic extension of prime degree. Since L i−1 contains therequisite roots of unity, L i is a simple radical or a simple pseudo radicalextension of L i−1 . Then M/k is a generalized radical extension and ourlemma is proved.We are, now, ready to proveTheorem 16. A separable polynomial f (x)∈k[x] is solvable by generalizedradicals if and only if its group is solvable.Proof. Let K be the splitting field of f (x) and G the galois group ofK/k. Suppose f (x) is solvable by generalized radicals. Then K ⊂ Lwhere L/k is galois and by lemma 1, has solvable galois group H. LetG o be the galois group of L/K Then H/G o is isomorphic to G and so Gis solvable.□Let, conversely, G be solvable. Then, by lemma 2, K is containedin a generalized radical extension and so f (x) is solvable by generalizedradicals.We can easily prove 159□

138 6. Special algebraic extensionsCorollary. A separable polynomial f (x)∈k[x] is solvable by radicalsif and only if its splitting field has a solvable galois group of order primeto the characteristic of k, if different from zero.Let k be a field. The polynomialf (x)= x m − x 1 x m−1 + x 2 x m−2···+(−1) m x m ,where x 1 ,..., x m are algebraically independent over k, is said to be thegeneral polynomial of the mth degree over k. It is so called, becauseevery monic polynomial of degree m over k is obtained by specializingthe values of x 1 ,..., x m to be in k. Let L=k(x 1 ,..., x m ). Let y 1 ,...,y mbe roots of f (x) over L. Thenf (x)=(x−y 1 )···(x−y m )and y 1 ,...,y m are distinct. The splitting field k(y 1 ,...,y m ) of f (x) overL is a galois extension whose galois group is isomorphic to S m . Hence,the general polynomial of the mth degree over k has a group isomorphicto the symmetric group on m symbols.But,S m is not solvable, if m>4, so that in virtue of theorem 16, wehave the theorem of Abel.Theorem 17. The group of the general polynomial f (x) of the mth degreeis isomorphic to the symmetric group S m on m symbols and hence,for m>4, f (x) is not solvable by generalized radicals.160We shall now explicitly show how to obtain the roots of a polynomialof degree≤4 in terms of generalized radicals.Let f (x) be a general polynomial of degree m over k and let K bethe splitting field. K/k has the group S m . Let y 1 ,...,y m be the roots off (x). Put ∏D= (y i − y j ) 2 .i< jThen D is fixed under all permutations in S m and, hence, D∈k.If we assume that the characteristic of k is 2, then k( √ D) is a galoisextension of k and K/k( √ D) has the galois group isomorphic to

8. Solvable extensions 139the alternating group on m symbols. D called the discriminant of thepolynomial f (x).Let us, first, consider the general polynomial of the second degreeThenf (x)=(x−y 1 )(x−y 2 )= x 2 − x 1 x+ x 2 .y 1 + y 2 = x 1 , y 1 y 2 = x 2 .Suppose, now, that k has characteristic2. ThenD=(y 1 − y 2 ) 2 = (y 1 + y 2 ) 2 − 4y 1 y 2 = x 2 1 − 4x 2.Also, y 1 + y 2 = x 1 , y 1 − y 2 =± √ D so thaty 1 = x 1+ √ D, y 2 = x 1− √ D22and K= k( √ D) is the splitting field of f (x) and is a radical extension.Let, now, k have characteristic 2. Then x 1 o, since f (x) is separable.Put x 1 x instead of x. Then the polynomial x 2 − x+ x 2x 2 has roots y 1x11and y 2. But this is a normal polynomial so that ifλ= x 2, then 161x 1 x 1 2y 1 = x 1λP , y 2=x 1λP + x 1and thus k( λ ) is a pseudo radical extension and is splitting field ofPf (x).We shall now study cubic and biquadratic polynomials.Let, first, k have characteristic2or 3. Let m=3 or 4 andf (x)= x m − x 1 x m−1 +···+(−1) m x mbe the polynomial of the mth degree whose roots are y 1 ,...,y m .If we put x+ x 1instead of x, we get a polynomial whose roots aremy 1 − x 1m ,...,y m− x 1m and which lacks the terms in xm−1 .

140 6. Special algebraic extensionsWe shall, therefore, take the polynomial f (x) in the formf (x)= x m + x 2 x m−2 +...+(−1) m x m .If y 1 ,...,y m are the roots, theny 1 + y 2 +...+y m = o.Also , D m = ∏ i − y j )i< j(y 2 . A simple computation shown thatandD 3 =−4x 3 2 − 27x2 3D 4 = 16x 4 2 x 4− 4x 3 2 x2 3 − 128x2 2 x2 4 + 144x 2x 2 3 x 4−27x 4 3 + 256x3 4 .If K is the splitting field of f (x) over L=k(x 1 ,..., x m ) then L( √ D)is the fixed field of the alternating group A m and L( √ D)/L is a radicalextension. In order to study the extension K/L( √ D), let us, first,take the case m=3. The symmetric group on 3 symbols, S 3 , has thecomposition seriesS 3 ⊃ A 3 ⊃ (e).162K/L( √ D) is thus a cyclic extension of degree 3. Letρbe a primitivecube root of unity and let M=L( √ D,ρ). Let N=KM be a compositeof K and M. Then KM/M has degree 1 or 3, according asρis in K ornot. In the first case, M=K. In the second case, KM/M is a cyclicextension of degree 3 over K and M contains the cube roots of unity.Thus KM=M( 3√ ω), for someω∈ M. In order of determine thisω, weuse Lagrange’s method.KM is the splitting field over M of the polynomial x 3 + x 2 x− x 3 . Lety 1 , y 2 , y 3 be roots of this polynomial. Putω=y 1 +ρy 2 +ρ 2 y 3ω ′ = y 1 +ρ 2 y 2 +ρy 3 .

8. Solvable extensions 141Thenω 3 = 27 2 x 3+ 3 √ D(ρ− 1 2 ). Changingρintoρ2 we getω ′3 .Henceω=ρ a 3 √272 x 3+ 3 √ D(ρ− 1 2 )and we have a similar expression forω ′ . Here a≥o. In order to determinea, we use the fact thatωω ′ =−3x 2 and so, choosing the rootof unityρ a forωarbitrarily, the root of unity in the expression forω ′ isuniquely determine. Nowy 1 + y 2 + y 3 = oy 1 +ρy 2 +ρ 2 y 3 =ωy 1 +ρ 2 y 2 +ρy 3 =ω ′and since the matrix⎞1 1 11 ρ ρ⎛⎜⎝2⎟⎠1 ρ 2 ρis non-singular, the values of y 1 , y 2 , y 3 are uniquely determined K·M is 163a radical of L and contains K.Consider, now, the polynomial of the fourth degreef (x)= x 4 + x 2 x 2 − x 3 x+ x 4whose roots are y 1 , y 2 , y 3 , y 4 with y 1 + y 2 + y 3 + y 4 = o. The galois groupof the splitting field K/k is S 4 . This has the composition seriesS 4 ⊃ A 4 ⊃ B 4 ⊃ C 4 ⊃ (e).Let K 1 , K b , K c be the fixed fields of A 4 , B 4 and C 4 respectively. NowA 4 is the alternating group, B 4 the group consisting of the permutations(1), (12)(34), (13)(24), (14)(23)and C 4 is the group of order 2 formed by(1), (12)(34).

142 6. Special algebraic extensionsK a = k( √ D) and is of degree 2 over k. Now K b /K a is of degree 3and is cyclic. Hence K b = K a (θ) whereθ∈Kis an element fixed by B 4but not by A 4 . Such as elements, for instance, isθ 1 = (y 1 + y 2 )(y 3 + y 4 ).θ 1 has 3 conjugatesθ 1 ,θ 2 ,θ 3 obtained fromθ 1 by operating onθ 1 , byrepresentatives of cosets of A 4 /B 4 . Thusθ 2 = (y 1 + y 3 )(y 2 + y 4 )θ 3 = (y 1 + y 4 )(y 2 + y 3 ).164Consider the polynomialϕ(x)=(x−θ 1 )(x−θ 2 )(x−θ 3 ).It is fixed under A 4 and so, its coefficients are in K a . A simple computationshows thatϕ(x)= x 3 − 2x 2 x 2 + (x 2 2 − 4x 4)x+ x 2 3 .ϕ(x) is called the reducing cubic or the cubic resolvent. By themethod adopted for the solution of the cubic, ifρis a primitive cuberoot of unity and M = K a (ρ), then K b M, the composite, is a radicalextension of k in whichθ 1 ,θ 2 ,θ 3 lie.K c /K b is of degree 2 and so K c = K b (α) whereαis fixed under B 4 ,but not by C 4 . such as element isα=y 1 + y 2 .Nowα 2 = (y 1 + y 2 ) 2 =−(y 1 + y 2 )(y 3 + y 4 )=−θ 1 . Thus K c =K b ( √ −θ 1 ). Hence, if K b M=K d , then K d (α) is a radical extension of kcontainingα. Put nowT= K d (α,β)whereβ=y 1 +y 3 = √ −θ 3 . Then T is a radical extension of k containingK. The indeterminacy signs in taking the square roots of−θ 1 and−θ 3can be fixed by observing that(y 1 + y 2 )(y 1 + y 3 )(y 1 + y 4 )= x 3 .

8. Solvable extensions 143we now havey 1 + y 2 = √ −θ 1 , y 3 + y 4 =− √ −θ 1y 1 + y 3 = √ −θ 2 , y 2 + y 4 =− √ −θ 2y 1 + y 4 = √ −θ 3 , y 2 + y 3 =− √ −θ 3 .for which y 1 , y 2 , y 3 , y 4 can be obtained. We have, hence, proved 165Theorem 18. If K has characteristic 2 or 3, then the cubic and biquadraticpolynomials over k can be radicals.Let us, now, assume that k has characteristic 3. Let x 3 + x 1 x 2 + x 2 x+x 3 be a cubic polynomial and K, its splitting field. K/k has the galoisgroup S 3 . Let L be the fixed field. K/k has the galois group S 3 . Let Lbe the fixed field of A 3 . ThenwhereL=k( √ D)D=(y 1 − y 2 ) 2 (y 2 − y 3 ) 2 (y 3 − y 1 ) 2 .K/L is now a cyclic extension of degree 3 and since k has characteristic3, K= L(ω), whereω 3 −ω−α=ofor someα∈L. Thus K is a generalized radical extension. We shallnow determineωandαand therefrom, y 1 , y 2 and y 3 .In order to do this, we have to consider two cases, x 1 = o, andx 1 o.Let, first, x 1 = o. Then y 1 + y 2 + y 3 = o. Letσbe a generator of thegalois group fo K/L and let notation be so chosen thaty σ 1 = y 2, y σ 2 = y 3, y σ 3 = y 1.Since S K/L y 1 = y 1 + y σ 1 + yσ2 1= o, by Hilbert’s theorem, there is aλin K such thaty 1 =λ σ −λ.

144 6. Special algebraic extensionsAlso 166∑x 2 = y 1 y 2 + y 2 y 3 + y 3 y 1 = (λ σ −λ)(λ σ2 −λ σ )=−(λ+λ σ +λ σ2 ) 2 .σSince x 2 o (otherwise, polynomials is not separable), we see thatx 2 =−t 2 , for some t∈L. The polynomial, therefore, has the formx 3 − t 2 x+ x 3 .Put now tx for x. Then the polynomial x 3 − x+ x 3 /t 3 has roots y 1t ,y 2t , y 3t . Letω= x 3/t 3 . Theny 1 = t ω P , y 2= t ω P + t, y 3= t ω P + 2tand thus the roots are all obtained.The indeterminacy in the sign of t does not cause any difficulty; for,if we use (−t) instead of t, then, observing that− ω P =−ω , we see thatPy 1 =−t −ωP , y 3=−t −ωP − t, y 2=−t −ωP + tso that y 2 and y 3 get interchanged.We now consider the case x 1 o.Put x+a instead of x. Then the new polynomial isx 3 + x 1 x 2 + x(x 2 + 2x 1 a)+ x 3 + x 2 a+a 2 + a 3 .167Choose a so that x 2 + 2x 1 a=o. For this value of a,µ= x 3 + x 2 a+a 2 + a 3 o; for, otherwise, the polynomial will be reducible and theroots are o, o,−x 1 . The roots of this new polynomial are y 1 − a, y 2 − a,y 3 − a.Let now 1 x be written for x, then the polynomial is reduced to x3 +x 1µ x+ 1 µ . We are in the previous case. The roots, now, are 1y 1 − a , 1y 2 − a ,1y 3 − a . If−t2 = x 11, t∈L andν=µ µt 3 , then K=L(ν ) and the rootsP

8. Solvable extensions 145y 1 , y 2 , y 3 are given byy 1 = a+ 1t ν , y 2 = a+P11t+t ν , y 3 = a+P2t+ t ν PSince the characteristic is 3, the biquadratic polynomial can be takenin the formx 4 + x 2 x 2 + x 3 x+ x 4 .The proof is similar to the old one( except that K b /K a is a cyclicω)extension of degree 3 and so K b = K a for a suitableωin K a . ToPfindω, we use the foregoing method. One finds that K is a generalizedradical extension.We now consider the case where k has characteristic 2. In this case,the cubic polynomial cab be taken in the form, x 3 + x 2 x+ x 3 . Let y 1 , y 2 ,y 3 be the roots. Theny 1 + y 2 + y 3 = oand therefore y 2 1 + y2 2 = y2 3and so on. PutThenω= y 1y 2+ y 2y 3+ y 3y 1,ω ′ = y 2y 1+ y 3y 2+ y 1y 3.ω+ω ′ = y 1(y 2 2 + y2 3 )+y 2(y 2 3 + y2 1 )+y 2(y 2 1 + y2 2 )y 1 y 2 y 3= 1,which shows thatωk, because ,then, it will be symmetric and equaltoω ′ . Thus k(ω) is a quadratic subfield of K. A simple computationshows thatωandω ′ =ω+1 are roots ofx 2 − x+ x 3 2 .K/k(ω) is cyclic of degree 3 and one uses the method of Lagrange 168to obtain a generalized radical extension.

146 6. Special algebraic extensionsSuppose now that f (x) is a polynomial of degree 4. Let f (x) =x 4 + x 1 x 3 + x 2 x 2 + x 3 x+ x 4 . We have to consider two cases. Let, first,x 1 = o. Then the roots y 1 , y 2 , y 3 , y 4 satisfyAs before, puty 1 + y 2 + y 3 + y 4 = o.θ 1 = (y 1 + y 2 )(y 3 + y 4 ), θ 2 =...,θ 3 =···The reducing cubic is thenϕ(x)= x 3 + x 2 2 x+ x 3.Furthermoreθ 1 +θ 2 +θ 3 = o. Put nowω= θ 1θ 2+ θ 2θ 3+ θ 3θ 1.Thenωis fixed under A 4 but not under S 4 . Also K a = k(ω) is asimple pseudo radical extension.Now K b /K a is a cyclic extension of degree 3 and has to be solvedby Lagrange’s methods. Further, K c /K b is of degree 2. Put nowThenω 1 = y 3y 1 y 2 y 4, ω ′ 1 = y 4y 1 y 2 y 3.ω 1 +ω ′ 1 = y2 3 + y2 4y 1 y 2 y 3 y 4= (y 1+ y 2 )(y 3 + y 4 )y 1 y 2 y 3 y 4= θ 1x 4.We assume x 3 o. Thenθ 1 o. If we putω 2 = x 4ω 1θ 1, ω ′ 1 = x 4ω ′ 1θ 1,169thenω 2 +ω ′ 2 = 1 and K c=K b (ω 2 ).In similar manner, K= K c (ω 3 ) whereω 3 = x 4θ 2,y 1y 2 y 3 y 4.

8. Solvable extensions 147Suppose, now, that x 1 o. Then, as before, we construct the fieldK b . In order to exhibit K c as a pseudo radical extension of K b , observethat y 1 + y 2 is fixed under C 4 but not under B 4 . Also(y 1 + y 2 ) 2 = (y 1 + y 2 )(y 3 + y 4 + x 1 )=θ 1 + x 1 (y 1 + y 2 )which shows that ( )y1 + y 2K c = K b .x 1( )y1 + y 3Similarly, K= K c .x 1Our contentions are completely established.

Chapter 7Formally real fields1 Ordered ringsA commutative ring R is said to be ordered if there is an ordering relation> (greater than) such that170(1) for every a∈R, a>o, a=o or−a>o.(1) a, b∈R, a>o, b>o⇒a+b>o, ab>o.We may then define a>b by a−b>o. If a>b, then, for any c inR, a+c>b+c and if c>o, ac>bc.If−a > o, we shall say a is negative and if a > o, a said to bepositive. We denote “a is negative” by a

150 7. Formally real fields171non-negative elements of R. The only element which is both positiveand negative is zero. Clearly, if R is a any subset P of R satisfying thefour conditions above again determine an order on R.Two elements a, b∈R, aob are said to have the same oropposite signs according as ab>oor ab o for ao in R. Moregenerally, every finite sum of squares of elements of R is positive. Theseelements will be contained in the set of non-negative elements in everyorder of R.Since R has a unit element 1 and 1 2 = 1, we have 1>o. Also,n·1=1+1···+1, n times so n>o. This proves1) An ordered ring with unit element has characteristic o.Let ao, bo be elements of R. Then a or−a is positive.Similarly b or−b is positive. Hence ab or−ab is positive whichproves that abo. Therefore2) An order ring is an integrity domain.We define an ordered field to be an ordered ring whose non-zeroelements form a multiplicative commutative group. We have3) If k is an ordered field, its positive elements form a multiplicativegroup.For, if x∈k, x>o; then xx −1 > o. If−x −1 > o, then−xx −1 > owhich contradicts x −1 x>o. Thus x −1 > o and (3) is proved.Suppose R and R ′ are two rings, R⊂R ′ . If R ′ is ordered, clearlyR is ordered by means of the induced order. If however, R isordered, it may not be possible, in all cases, to extend this orderto R ′ . However, in case, it is always possible, namely4) If K is the quotient field of an ordered ring R then the order in Rcan be uniquely extended to K.172Proof. Let R have an order. It is an integrity domain. Any elementsx∈Kis of the form x= a , a, b∈R, bo. Let xo. Then ao.b

1. Ordered rings 151define x>o by ab>o. Then this defines an order in K. In the firstplace, the definition does not depend on the way x is expressed in theform a a′. Suppose x=b b ′ . Then ab′ = a ′ b. Since b ′ o, multiplyingboth sides of this equality by bb ′ , we haveab·b ′2 = a ′ b ′· b 2 .Since ab>o, b ′2 > o, b 2 > o, it follows that a ′ b ′ > o, that is a′b ′> o.In order to prove that A 1 ,..., A 4 are satisfies, let x= a b > o andy= a′b ′> o. Then ab>oand a′ b ′ > o.x+y= ab′ + a ′ bbb ′Now (ab ′ + a ′ b)bb ′ = ab·b ′2 + a ′ b ′· b 2 . Since ab>o, a ′ b ′ > o,b 2 > o, b ′2 > o, it follows that (ab ′ + a ′ b)bb ′ > o or x+y>o. In similarmanner, xy>o.Suppose x≥o and y≥o and x+y=o; then x=o, y=o. For, ifx= a b , y= c ad+ bc, then ab≥o, cd≥o and x+y=d bd□= o, so thatad+ bc=o. Thus abd 2 + cdb 2 = o, which means that since all elementsare in R, ab=o, cd=o, i.e. a=o=c.If x∈R, than x= ab a . If x=o, then ba2 = o or b=o, so that theorder coincides on R with the given order in R.That the extension is unique can be, trivially, seen.Since every ordered field has characteristic zero, it contains a sub- 173field isomorphic toΓ, the rational number fieldΓhas thus induced order.We shall now prove(5) Γ can be ordered in one way only.For, if Z denotes the set of integers, thenΓis the quotient field of Z.On Z, there is only one order since 1>o and hence n=1+1+···+1>o.Thus, all natural integers have to be positive.

152 7. Formally real fields2 Extensions of orders174Hereafter, we consider ordered fields k. Our main task will be the studyof extensions of orders in k to extension fields K of k. For this purpose,we introduce the notion of a positive form on k.∑Let k be an ordered field. A polynomial m a i x 2 i , a i∈ k, is said to bean m-ary form over k. It is said to be positive if a i > o, i=1,...m. Anm-ary form is said to represent a∈k, if there existα 1 ,...,α m in k suchthat ∑a i α 2 i= a.iClearly a positive form represents zero, only ifα 1 ,...,α m = o. Letk be an ordered field and K/k, an extension field. We shall proveTheorem 1. K has an order extending the order in k, if and only if,every positive form over k is still positive in K.Proof. We have only to prove the sufficient. To this end, consider thefamily M of subsets{M α } of K having the following properties. Denote,by S , the set of elements in K of the form∑a i α 2 i ,a i ∈ k and a i > o andα i ∈ K. (α i can be all zero also).ii=1□Then1) M α ⊃ S 2) M α + M α ⊂ M α3) M α M α ⊂ M α 4) M α ∩ (−M α )=(o).This family is not empty, since S satisfies this condition. In the usualway, we make M a partially ordered set and obtain a maximal set P. Wehave now to show thatP∪(−P)= Kand then P will determine an order. Let xo be an elements of K whichis not in P. DefineQ=P− xP

2. Extensions of orders 153as the set of elements of the form a− xb, a, b∈P. Obviously Q satisfies(1). To see that Q satisfies (2), observe that if a− xb, c− xd are in Qthen(a− xb)+(c− xd)=(a+c)− x(b+d);but a, b, c, d being in P which is an element of M, a+c, b+d arein P. In a similar way Q satisfies (3). That Q satisfies (4) can be seenas follows: Let a− xb and c− xd be in Q with a, b, c, d, in P. Let(a+c)− x(b+d)=o. Then b+d=o. For, if b+do, thenx= a+cb+d = (a+c)(b+d) 1(b+d) 2and so is an element of P, which is a contradiction. Hence b+d=0 175and, therefore, a+c=0. Since a, b, c, d are all in P, it follows thata=b=c=d=0. Thus Q∈ M. But Q⊃P so that, by maximality ofP, Q=P. This means that−x∈P or x∈−P. The theorem is thereforeproved.IfΓis the field of rational numbers and n is a positive integer, thenn=1+1+···+1. If r= a is a positive rational number, thenbab = ab 1+1···+1b2= b 2 ,so that every positive rational number is a sum of squares. This showsthat every positive form overΓcan be put in the form x 2 1 +···+ x2 n .If k is an ordered field then ∑ α 2 i= 0,α i ∈ k implies thatα i = 0,ii=1,.... On the other hand, if k has the property that ∑ α 2 i= 0,α i ∈ kimpliesα i = 0, then n=1+1+···+10 so that k has characteristic zero.Since every positive form overΓis essentially of the type x 2 1 +···+ x2 n,we have theTheorem 2. k is ordered if and only if ∑ ii=1, 2,....α 2 iα 2 i= 0,α i = k impliesα i = 0,It is obvious that, if, in a field k, ∑ = 0, withα 1 ,α 2 ,... not allizero, then ∑−1= β i 2,i

154 7. Formally real fields176β i ∈ k. A field in which−1 is not a sum of squares is called a formallyreal field. From theorem 2, it follows that formally real fields areidentical with ordered fields.We shall, now, prove the following application of theorem 1.Theorem 3. Let k be a formally real field with a given order and f (x),an irreducible polynomial in k[x]. Let a, b be two elements in k such thatf (a) f (b) 0 in k. This means that, in k[x],∑1+ a i {ϕ i (x)} 2 = f (x)ψ(x).iiSince f (x) has degree n and left side has degree≤ 2n−2,ψ(x) has,at most, the degree n−2. □We now use induction on n. If n=1, these is nothing to prove.Assume theorem proved for n−1 instead of n. Let g(x) be in irreduciblefactor ofψ(x). Then g(x) has degree≤n−2. Now∑0

2. Extensions of orders 155Ifβis a root of g(x) inΩ, then g(β) = f (β) = 0. But by inductionhypothesis, k(β) has an order extending the given order in k. Hence∑0=1+ a i {ϕ i (β)} 2 > 0,iwhich is a contradiction. Thus our theorem is completely proved.Suppose k is an ordered field, let us denote, by|a|, the absolute valueof a∈kby⎧0 if a=0⎪⎨|a|= a if a>0⎪⎩ −a if a M, thent −n f (t)=1+a 1 t −1 +···+a n t −n > 0which shows that t n and f (t) have the same sign.Suppose now f (x) is irreducible and of odd degree. Then, if M isdefined as above,f (M) f (−M)0, then the polynomial x 2 − a changes sign in k. 178For, (a+1) 2 > a and−a

156 7. Formally real fields3 Real closed fieldsWe had seen above that, under certain circumstances, an order in a fieldk can be extended to a finite algebraic extension of k. We shall consider,now, a class of fields called real closed fields defined as follows: k issaid to be real closed, if1) k is ordered2) k has no proper algebraic extension K with an order extending thatin k.Before we establish the existence of such fields, we shall obtainsome of their properties. We first proveTheorem 4. For a formally real field k, the following properties areequivalent:1) k(i) is algebraically closed, i being a root of x 2 + 1.2) k is real closed.3) Every polynomial of odd degree over k has a root in k and everypositive element of k is a square in k.179Proof. 1⇒2. k(i) being of degree 2 over k, there are no intermediaryfields, so that, k being ordered, and k(i) being algebraically closed, k hasno ordered algebraic extension.2 ⇒ 3. Suppose f (x) is a polynomial of odd degree. Then itchanges sign in k. Hence an irreducible factor of f (x), also of odd degree,changes sign in k. Ifαis a root of this irreducible factor, then k(α)is ordered with an order extending that in k. Henceα∈k. If a>0 in k,then x 2 − a changes sign in k. Thus √ a∈k.3⇒1. The poof of this part consists of three steps. Firstly, everyelement of K= k(i) is a square. For, let a+bi be an element of K, a,b∈k. Puta+ib=(c+id) 2

3. Real closed fields 157where c and d have to be determined in k. Since 1, i form a base of K/kwe getc 2 − d 2 = a, 2cd= b.Therefore, (c 2 + d 2 ) 2 = a 2 + b 2 . But a 2 + b 2 > 0 in k and, since everypositive element is a square, there is aλ>0 in k such thatSolving for c 2 and d 2 , we havec 2 + d 2 =λ.c 2 = λ+a2 , d2 = λ−a2 .But, sinceλ 2 = a 2 + b 2 , it follows thatλ≥|a|,λ ≥|b|. Thereforeλ+a≥ 0, λ−a ≥ 0. Therefore2 2√ √λ+a λ−ac=±2 , d=± 2exist in k. The arbitrariness in the signs of c and d can be fixed from thefact that 2 cd= b.□This proves that every quadratic polynomial over k has a root in 180K. For, if ax 2 + bx+c ∈ k[x], then, in an algebraic closure of K,−b± √ b 2 − 4acare its roots (a0). But √ b2a2 − 4ac∈K.The second step consists in showing that every polynomial in k[x]has a root in K. Let f (x) be in k[x] and let N be its degree N= 2 n· q, qodd. We shall use induction on n. If n=0, then N= q, and so whateverodd number q be, f (x) has a root already in k. Let us, therefore, assumeproved that every polynomial of degree 2 n−1 q ′ , where q ′ is odd, withcoefficients in k has a root in K. Let f (x) be of degree N= 2 n· q, q odd.Letα 1 ,...,α t be the distinct roots of f (x) in an algebraic closure of K.Letµ∈kbe an element to be suitably chosen later. Putλ i j (µ)=λ i j =α i +α j +µα i α j i, j=1,...,t, i j.

158 7. Formally real fieldsConsider now the polynomial∏ϕ µ (x)= (x−λ i j ).i j181N(N− 1)This has degree = 2 n−1· q ′ , q ′ odd. Also, by every permutationof the symbols 1,...,t, the polynomial goes over into itself.2Thusϕ µ (x)∈k[x]. Since its degree satisfies the induction hypothesis,for everyµ∈k, there is an i and a j such thatλ i j (µ)∈K. Since k isan infinite field, there existµ,µ ′ ,µµ ′ and both in k such thatλ i j (µ)andλ i j (µ ′ ) for two integers i and j are in K. This means thatα i α j and,hence,α i +α j are in K. The polynomial x 2 − x(α i +α j )+α i α j is apolynomial in K[x]. By what we proved above, both its roots are in K.Thus our contention is proved.The third step consists in proving that every polynomial in K[x],has a root in K. For, let f (x) be a polynomial in K[x]. Letσbe thegenerating automorphism of K/k. It is of order 2. Denote by f σ (x)the polynomial obtained from f by applyingσ on the coefficients of f .Thenϕ= f (x) f σ (x) is a polynomial in k[x]. The second step shows thatϕ has a rootαin K. Furthermore ifαis a root ofϕ,α σ is also a root ofϕ, so that eitherα is a root of f (x) orα σ is a root of f (x).We have thus proved theorem 4 completely.We deduce from this an important corollary due to Artin andSchreier.Corollary 1. IfΩis an algebraically closed field and K a subfield suchthat 1

3. Real closed fields 159Hence every sum of two squares and, hence, of any number ofsquares is a square. Therefore∑−1=iis impossible in K. By theorem 2, therefore, K is formally real. □ 182This proves that the real closed fields are those and only those whichare such that their algebraic closures are finite over them.We have againCorollary 2. A real closed field has only one order.For, the set of positive elements coincides with the set of squares ofthe elements of the field.This shows that, if on ordered field has two distinct orders, it hasalgebraic extensions which are ordered. It must be remembered that ifa field has only one order, it is not necessarily real closed. The rationalnumber field, for example, has only one order.Suppose k is a real closed field. Then every irreducible polynomialin k[x] is of degree one or two. Suppose f (x) is a polynomial in k[x] anda, b in k such thatf (a) f (b)

160 7. Formally real fieldsAll the roots of f (x) that lie in k lie between±M.Let k be a real closed field and f (x) a polynomial in k[x]. Let f ′ (x)be its derivative. Putϕ 0 = f ,ϕ 1 = f ′ . Since k[x] is a Euclidean ring,define, by the Euclidean algorithm, the polynomialsϕ 0 = A 1 ϕ 1 −−ϕ 2ϕ 1 = A 2 ϕ 2 −−ϕ 3... ...ϕ r−1 = A r ϕ r .It is then well-known thatϕ r (x) is the greatest common divisor ofϕ 0andϕ 1 . The sequenceϕ 0 ,ϕ 1 ,...,ϕ r of polynomials in k[x], is known asthe Sturmian polynomial sequence.Let a∈k be such thatϕ 0 (a)0. Thenφ r (a)0. Consider theset of elementsϕ 0 (a),ϕ 1 (a),...,ϕ r (a) in k. The non-zero ones amongthem have a sign. Denote, byω(a), the number of changes of sign in thesequence of elements,ϕ 0 (a),ϕ 1 (a),...,ϕ r (a),in this order, taking only the non-zero elements. A very important theoremdue to Sturm is184Theorem 6. Let b and c be two elements of k, b

3. Real closed fields 161so that (x−α) t−1 is the highest power of of x−α, that dividesϕ 1 (x).Hence¯ϕ 0 (x)=(x−α)ψ 2 (x)¯ϕ 1 (x)=tψ 2 (x)+(x−α)ψ 3 (x),ψ 2 (x) andψ 3 (x) being polynomials over k. Note thatϕ¯1 (x) is not thederivative of ¯ϕ 0 (x). We shall drop the ‘bars’ on theϕ ′ s and write themasϕ 0 ,ϕ 1 ,...,ϕ r−1 ,ϕ r = 1 andϕ 0 having no multiple roots. Note thatω(b) orω(c) is not altered by doing the above. □The finite number of polynomialsϕ 0 ,ϕ 1 ,...,ϕ r−1 have only finitelymany roots between b and c. By means of these roots, we shall split theinterval (b, c) into finitely many subintervals, the end points of whichare these roots. We shall study how the functionω(a) changes as a runsfrom b to c.1) No two consecutive functions of the Sturmian series ϕ 0 (x), 185ϕ 1 (x),...,ϕ r−1 (x) can vanish at one and the same point, inside theinterval (b, c). For, suppose b

162 7. Formally real fieldsis true ofϕ l+1 . Thus, whatever signϕ l might have in L and R, thefunctionω(a) remains constant when a goes from L to R crossing azero ofϕ l , 0

3. Real closed fields 163187Proof. 1) Let f (x) be a polynomial in k[x] andα 1 ,...,α t the distinctroots of f (x) in K. Letα ∗ 1 ,...,α∗ t be the distinct roots of f (x) inK ∗ . Put L = k(α 1 ,...,α t ) and L ∗ = k(α ∗ 1 ,...,α∗ t ). Since L/k isfinite, L = k(ξ) for someξin K. Supposeϕ(x) is the minimumpolynomial ofξ in k[x]. Letξ ∗ be a root ofϕ(x) in K ∗ . Then k(ξ) andk(ξ ∗ ) are k-isomorphic. Ifρis this isomorphism, thenρξ=ξ ∗ . Butk(ξ)= L=k(α 1 ,...,α t ). Henceρα 1 ,...,ρα t will be distinct rootsof f (x) in K ∗ . Thus L ∗ = k(ξ ∗ ). HenceL≃ L ∗ .2) Supposeϕ(x) is any polynomial in k[x],β 1 ,...,β s its distinct rootsin K. Letβ ∗ 1 ,...,β∗ s be the corresponding roots in K∗ . Consider allthe positive quantities amongβ i −β j , i j. Their square roots existin K. Letψ(x) be a polynomial in k[x], among whose roots are thesesquare roots and letδ 1 ,...,δ g be the roots ofψ(x) in K,δ ∗ 1 ,...,δ∗ gthe corresponding roots in K ∗ . Then, from above,F= k(β 1 ,...,β s , δ 1 ,...,δ g )≃≃ k(β ∗ 1 ,...,β∗ s , δ∗ 1 ,...,δ∗ g )=F∗ .Letτbe this isomorphism. Let notation be such thatτ(β i )=β ∗ iτ(δ i )=δ ∗ i .Supposeβ i >β j . Thenβ i −β j > 0 so thatβ i −β j =δ 2 t , for some t.Also,τ(β i −β j )=β ∗ i−β ∗ j . Butτ(β i −β j )=τ(δ 2 t )=δ ∗2t .Henceβ ∗ i−β ∗ j =δ∗2 t> 0, which proves thatβ ∗ i>β ∗ j .The isomorphismτ between F and F ∗ preserves order between the 188roots ofϕ(x) in K.

164 7. Formally real fields3) In order, now, to construct the isomorphism between K and K ∗ , weremark that any such isomorphism has to preserve order. For, ifσisthis isomorphism andα>βin K, thenα−β=δ 2 so thatσ(α−β)=σ(δ 2 )=(σ(δ)) 2 > 0so thatσα>σβ. We shall, therefore, construct an order preservingmap which we shall show to be an isomorphism.4) Letαbe an element in K, h(x) its minimum polynomial over k. Letα 1 ,...,α t be the distinct roots of h(x) in K and let notation be sochosen thatα,

3. Real closed fields 165Theorem 8. If k is an ordered field with a given order andΩ, its algebraicclosure, there exists inΩ, upto k-isomorphism, only one realclosed field K with an order extending the given order in k.Proof. Let V be the family of formally real subfields ofΩ/k which havean order extending that in k. V is not empty, since k∈V. We partiallyorder V by inclusion. Let{k α } be a totally ordered subfamily of V. LetK 0 = ⋃ k α . Then K 0 is a field which is contained in V. This can beαeasily seen. By Zorn’s lemma, there exists a maximal element K in V.K has an order extending the order in k. To prove that K is real closed, letf (x) be a polynomial of odd degree over K. It changes sign. Thereforethere is an irreducible factor, also of odd degree, which changes signin K. This factor must have a root in K. Else, by theorem 3 thereexists an algebraic extension with an order extending that in k. This willcontradict maximality of K. In a similar way, every positive element ofK is a square. By theorem 4, K is real closed. If K and K ∗ are two realclosed subfields with orders extending that in k, then, by theorem 7, they 190are k-isomorphic.□Suppose now that k is a perfect field andΩ, its algebraic closure.Let G be the galois group ofΩ/K. If G has elements of finite order (notequal to identity), let K be the fixed field of the cyclic group generatedby one of them. Then (Ω : K) is finite and by Artin-Schreier theoremthis order has to be two. Thus any element of finite order in G has tohave the order two. Furthermore, in this case, k is an ordered field.On the other hand, if k is an ordered field andΩ, its algebraic closure,there exists, then by theorem 8, a real closed subfield ofΩ/k, sayK. This means that G has an element of order 2. Moreover no two elementsof order 2 commute. For ifα,βare of order 2 and commute, then1,α,β,αβ is a group order 4 which must have a fixed field L such that(Ω : L)=4. This is impossible, by Artin-Schreier theorem. Hence theTheorem 9. If k is a perfect field,Ω its algebraic closure and G thegalois group ofΩ/k then G has elements of finite order if and only if kis formally real. Also, then, all these elements have order 2 and no twoof them commute.

166 7. Formally real fields4 Completion under an orderLet k be a formally real field with a given order.function|| on k with values in k such thatWe had defined a|ab|=|a|·|b|,|a+b|≤|a|+|b|.191This implies that|a|−|b|≤|a−b|.Also a→|a| is a homomorphism of k ∗ into the set of positive elementsof k. The function|| defines a metric on the field k. We definea Cauchy sequence in k to be a sequence (a 1 ,...,a n ,.) of elements of ksuch that for everyε>0 in k, there exists n 0 , an integer such thatObviously, if m>n 0 + 1,| a n − a m |n 0 .|a m |=|a m − a no − a no |0in k, there is an integer n 0 = n 0 (ε) such that| a n |n 0 .The sum and product of two Cauchy sequence is defined as follows:-(a 1 , a 2 ,...)+(b 1 , b 2 ,...)=(a 1 + b 1 , a 2 + b 2 ,...)(a 1 , a 2 ,...)−(b 1 , b 2 ,...)=(a 1 + b 1 , a 2 + b 2 ,...)and it is easy to verify that the Cauchy sequences in k form a ring R andthe null sequences, an ideal Y of R. We assert that Y is a maximal

4. Completion under an order 167ideal. For, let (a 1 ,...) be a Cauchy sequence in k which is not a nullsequence. Then there exists aλ>0 in k and an integer n such that| a m |>λ, m>n. (1)For, if not, for everyε>0 and integer n, there exist an infinity of 192m>n for which| a m | n 0 such that|a m0 |m 0 ,|a m |≤|a m − a m0 |+|a m0 < 2εwhich proves that (a 1 ,...) is a null sequence, contradicting our assumption.Let 1 be the unit element of k. Then the sequence (1, 1,...) is aCauchy sequence and is the unit element in R. Let c=(a 1 ,...) be inR but not in Y . Let m be defined as in (1). Then c 1 = (0, 0,...0, a −1m,a −1m+1,...) is also a Cauchy sequence. For, letε>0and n, an integer sothat| a n1 − a n2 | n 0 .Then ∣ ∣∣∣∣∣1a n1− 1a n2∣ ∣∣∣∣∣=1|a n1 ||a n2 |∣∣a n2 − a n2∣ ∣∣< ελ 2,if n 1 , n 2 > max(m, n 0 ). Also cc 1 = (0, 0, 0,...,1, 1,...) which provesthat cc 1 ≡ (1, 1, 1,... (modY ). This proves that Y is a maximal idealand, therefore,1) R/Y is a field ¯k.¯k is called the completion of k under the given order. For everya∈k, consider the Cauchy sequence ā=(a, a,...). Then a→ā is anon-trivial homomorphism of k into ¯k.Since ¯k is a field, this is an isomorphism. ¯k thus contains a subfield 193isomorphic to k. We shall identify it with k itself.

168 7. Formally real fields(2) ¯k is an extension field of k.We shall now make ¯k an ordered field with an order which is theextension of the order in k. To this end, define a sequence c=(a 1 ,...,)of R to be positive if there is anε>0 and an n 0 = n 0 (ε) such thata n >ε, n>n 0 .If b=(b 1 ,...) is a null sequence, thenIf p>max(m, n 0 ), then|b n |m=m(ε).a p + b p >ε/2which shows that a+b is also a positive sequence. The definition ofpositive sequence, therefore, depends only on the residue class mod Y .Let P denote the set of residue classes containing the positive sequencesand the null sequence. We shall show that P determines anorder in ¯kFirst, let c 1 and c 2 be two positive sequences. There existε 1 > 0,ε 2 > 0 and two integers n 1 and n 2 such thata p > E 1 , p>n 1 ,b p > E 2 , p>n 2 ;if p>max(n 1 , n 2 ), thena p + b p >ε 1 +ε 2 ,194so that c 1 + c 2 is a positive sequence or P+ P⊂P. In a similar manner,PP⊂P.Let now c=(a 1 ,...) be not a null sequence. Then c and−c cannotboth be positive. For then, there existε,ε ′ both positive and integers n,n ′ such that ⎧⎪ ⎨a p >ε, p>n⎪ ⎩−a p >ε ′ , p>n.

4. Completion under an order 169If p>max(n, n 1 ), then0=a p − a p >ε+ε ′ > 0,which is absurd. Thus P∩(−P)=(0).Suppose now that c=(a 1 ,...) is not a null sequence. Suppose itis not positive. Then−c=(−a 1 ,−a 2 ,...) is a positive sequence. For,otherwise, for everyεand every n,−a p n.But, c being not positive, we havea p n 1 .c being a Cauchy sequence, there exists n 0 withHence, if p>max(n 0 , n 1 ),|a p − a q |n 0 .|a p |≤|a p − a q |+|a q |0, there exists n 0 such that the elementā−(a n0 ) in k has all its elements, from some index on, less thanεinabsolute value. This justifies our notation. One clearly haslim a n + lim b n = lim a n + b n .

170 7. Formally real fieldslim a n . lim b n = lim a n + b n .lim a n= lim a n,lim b n b nif (b 1 , b 2 ,...) is not a null sequence.IfΓis the rational number field, it is ordered and the completion ¯Γunder this unique order is called the real number field.This method of construction of the real number field goes back toCantor.5 Archimedian ordered fieldsOrdered fields can be put into two classes.A field k is said to be archimedian ordered if, for every two elementsa, b in k, a>0, b>0 there exists an integer n such thatnb>a196and, similarly, there is an integer m with ma>b.We may state equivalently that, for every a>0, there is an integer nwithn>a.A field k is said to be non-archimedian ordered if there exists a>0such thata>nfor every integer n.Γ, the rational number field is archimedian ordered. Consider, now,the ringΓ[x] of polynomials. Forf (x)=a 0 x n + a 1 x n−1 +···+a n , a 0 0define f (x)>0, if a 0 > 0 as a rational number. That this is an order canbe verified easily. Also the order is non-archimedian becausex 2 + 1>n,

5. Archimedian ordered fields 171for every integer n. This order can be extended toΓ(x).This examples shows that an archimedian order in k can be extendedinto a non-archimedian order in an extension field which is transcendentalover k. That this is not possible in algebraic extensions is shown byTheorem 10. If k is archimedian ordered and K/k is algebraic withan order extending the order in k, the extended order in K is againarchimedian.Proof. Letαbe> 0 in the extended order in K. Let f (x)= x n +a 1 x n−1 +···a n be the minimum polynomial ofαin k[x]. Consider the quantities 1971−a i , i=1,...,n. They are in k and since k and is archimedian ordered,there exists an integer t>0 such thatt>1−a i , i=1,...,n.We now assert thatα

172 7. Formally real fields198In order to prove the inequality the other way round, let R be thering of Cauchy sequences in k and Y , the maximal ideal formed bynull sequences. We shall show that every residue class of R/Y can berepresented by a Cauchy sequence of rational numbers. Therefore, letc=(a 1 , a 2 ,...) be a positive Cauchy sequence in k. Since k is archimedianordered, there exists, for every n greater than a certain m, an integerλ n such thatλ n < na n < 1+λ nwhich means that∣ ∣∣∣∣a n − λ nn∣ < 1 n .Let d=(0, 0, 0,... λ mm ,λ m+1,...) be a sequence of rational numbers.Letε>0 be any positive quantity in k. There exists, then, anm+1integer t such thatεt>1, since k is archimedian ordered. Then, forn>tand m,∣∣a n − λ n∣∣

5. Archimedian ordered fields 173That such a sequence can be found is easily seen. Put now199c n = b n + c−b2 n .Then c 0 , c 1 ,... is a decreasing sequenceb≤b n ≤ b n+1 ≤ c n+1 ≤ c n ≤ c.Also (b n ) and (c n ) are Cauchy sequences; for,| b n+1 − b n |< c−b2 n+1.Furthermore, the sequence (c n − b n ) is a null sequence. For,c n = b n = c−b2 nThere is, therefore, anαbetween b and c such thatα=lim b n = lim c n .By definition ofλ n , f (c n )

Chapter 8Valuated fields1 ValuationsLet k be a field. A valuation on k is a function|| on k with valuesin the real number field satisfying200(1) | 0|= 0(2) | a|> 0, if a0(3) | ab|=| a|·|b|(4) | a+b|≤| a|+|b|where a and b are elements in k. It follows that a→| a| is a homomorphismsof k ∗ into the multiplicative group of positive real numbers. Ifwe denote by 1 the unit element of k, then| 1|=| 1 2 |=| 1| 2 =| 1|·|1|so that| 1|= 1. Ifζ is a root of unity, say an n say an n th root of unity,then1=|ρ n |=|ρ| nand so|ρ|= 1. Thus means that|−1|= 1 and so, for a∈k,|−a|=| a|.175

176 8. Valuated fieldsAlso, since a=a−b+b, we get| a|−|b|≤ a−b|.A valuation is said to be trivial if|a|= 1 for all a0.Two valuations|| 1 and|| 2 are said to be equivalent if for every a0in k,| a| 1 < 1⇒| a| 2 < 1,| a| 1 = 1⇒| a| 2 = 1.201It is obvious that the above relation between valuations is an equivalencerelation. All valuations equivalent to a given valuation form anequivalence class of valuations.If|| is a valuation then, for 0≤c≤1,| 1| c is also a valuation. Weshall now proveTheorem 1. If|| 1 and|| 2 are equivalent valuations, there exists a realnumber c>0 such that| a| 1 =| a| c 2for all a∈k.Proof. Let us assume that|| 1 is non-trivial. Then|| 2 is also non-trivial.Also, there exists a b∈ksuch that| b| 1 > 1,| b| 2 > 1. Let a∈k, a0.Then|a| 1 and|b| 1 being positive real numbers.| a| 1 =| b| λ 1whereλ= log|a| 1log|b| 1.□We approximate to the real numberλfrom below and from aboveby means of rational numbers. Let m | b| m/n1

2. Classification of valuations 177which means that|thata nb m| 1> 1. Since|| 1 and|| 2 are equivalent, this means| a| 2 >| b| m/n2.In a similar manner, if p/q>λ, then| a| 2 0, our theorem follows.2 Classification of valuationsA valuation is said to be archimedian if for every a∈k, there exists aninteger n=n(a) (that is ne, if e is the unit element of k) such that| a| 1.Proof. If|| is archimedian, there exists a∈k with| a|> 1 and aninteger n with| n|>| a|. This means that| n|> 1.□

178 8. Valuated fieldsLet|| be a valuation and n, a rational integer so that| n|> 1. Let a anyelement of k. If| a|≤ 1, then clearly| a| 1. Sincearchimedian axiom holds in real number fields, we have an integerm with| a|1. Then m=| n| λ(λ= log m > 0). Letµbe a positive integer greater thanλ. Thenlog|n|m

2. Classification of valuations 179Taking m th roots and making m→∞we get| a+b|≤ Max(| a|,| b|).The converse is trivial, since|n|=| 1+···+1|≤ 1.□We deduce easily 2045) If|| is non-archimedian and|a|| b|, then| a+b|= Max(| a|,| b|).Proof. Let, for instance,|a|>| b|. ThenAlso a=a+b−b, so that| a+b|≤ Max(| a|,| b|)=| a|| a|≤ Max(| a+b|,| b|).But, since| a|>| b|,| a+b|≥| b|. Thus| a|≤| a+b| and ourcontention is proved. □More generally, we have, if| a 1 |>| a j |, j1, then| a 1 + a 2 +...a n |=| a 1 |.In the case of non-archimedian valuations, many times, the so-calledexponential valuation is used. It is defined thus: If|| 0 is an non-archimedianvaluation, define the function|| by| a|=− log|a| 0 , a0.This has a meaning since| a| 0 > 0 for a0. We introduce a quantity∞ which has the property∞+∞=∞∞+a=∞

180 8. Valuated fieldsfor any real number a anda∞ = 0for any real number a. Then|| satisfies1’) | 0|=∞2’) | a| is a real number3’) | ab|=| a|+|b|4’) | a+b|≥ Min (| a|,| b|).205Then|1|=0,|ρ|=0for a root of unityρand the valuation is trivialif|a|=0 for a0. For two valuations|| 1 and|| 2 which are equivalent|a| 1 = c|a| 2 ,c〉0 being a real number.3 ExamplesFirst, letΓbe the finite field of q elements. Every element ofΓ ∗ satisfiesthe polynomial x q−1 − 1. Therefore, any valuation|| onΓis trivial.Let nowΓbe the field of rational numbers.Let|| be an archimedian valuation onΓ. It is enough to determine itseffect on the set of integers inΓ. There is an integer n such that|n|>1.Let m be any positive integer. Thenn=a 0 + a 1 m+···+a t m t ,[ ] log nwhere t≤ , 0≤a i < m. Therefore|a i |≤a i < m so thatlog m|n|≤m(t+1) max(1,|m| t ).

3. Examples 181Replace n by n r , where r is a positive integer. Then again, we haveMaking r→∞, we get1|n|≤m r (r log n 1log m + 1) r max(1,|m| t ).This proves that, since|n|>1,|n|≤max(1,|m| log nlog m).log n|n|≤|m| log mand|m|>1. 206Now we can repeat the argument with m and n interchanged andthus obtainlog n|m|≤|n| log mCombining the two inequalities we getlog|n|log n = log|m|log m .Since m is arbitrary, it follows that|m|=m Cwhere c>0 is a constant. Obviously, the valuation is determined byits effect on positive integers. From the definition of equivalence,|| isequivalent to the ordinary absolute value induced by the unique order inΓ.Let now|| be a non-trivial non-archimedian valuation. It is enoughto determine its effect on Z, the ring of integers.Since|| is non-trivial, consider the set Y of a∈Zwith|a|

182 8. Valuated fieldsAlso, if|a|

3. Examples 183208We shall denote the ordinary absolute valuation by|| ∞ .Let us consider the case of a function field K over a ground fieldk and let L be the algebraic closure of k in K. L is called the field ofconstants of the function field K. The valuations on K that we shallconsider shall always be such that|a|=ofor a∈k. (We consider exponential valuation). Hence valuations offunction fields are always non-archimedian, since the prime field is containedin k.Let, now,αbe in L. Thenαsatisfies the equationα n + a 1 α n−1 +···+a n = owhere a 1 ,...,a n ∈ k. If|| is a valuation of K then|α| n ≥ min(|α| n−1 ,...,|1|).From this, it follows that|α| ≥ 0. Also, since 1/α is algebraic,|α|≤0. Hence for allα∈L|α|=0.Thus the valuation is trivial on the field of constants.We shall consider the simple case where K= k(x), x transcendentalover k. Here L=k. Let|| be a valuation and let|x|

184 8. Valuated fieldsSuppose now that|x|≥0. As in the case of rational integers, considerthe subset Y of k[x] consisting of polynomials f (x) with| f (x)|>0. Then, since for everyφ(x) in k[x],|ϕ(x)|≥0, it follows that Y is amaximal ideal generated by an irreducible polynomial p(x). As in the 209case of the rational number field|R(x)|=λ|p(x)|where R(x)={p(x)} λ A(x) where A(x) and B(x) are prime to p(x) andB(x)λ is a rational integer. If we denote, by|| p(x) , this valuation and putλ=ord p(x) R(x), then|R(x)|=c ord p(x) R(x)where c=|p(x)| p(x) > 0. Every irreducible polynomial also gives rise toa valuation of this type. HenceTheorem 3. A complete system of inequivalent valuations of k(x) isgiven by the valuations induced by irreducible polynomials in k[x] andthe valuation given by the difference of degrees of numerator and denominatorof f (x) in k(x).4 Complete fieldsLet k be a field and|| a valuation of it. The valuation function defineson k a metric and one can complete k under this metric. The method isthe same as in the previous chapter and we give here the results withoutproofs.A sequence (a 1 , a 2 ,...) of elements of k is said to be a Cauchy sequenceif for everyε>0, there exists an integer n=n(ε) such that|a n1 − a n2 | n.It is a null sequence if for everyε>0, there is an integer n such that|a m |n.

4. Complete fields 185210The Cauchy sequences form a commutative ring R and the null sequencesa maximal ideal Y , therein. The quotient ¯k=R/Y is calledthe completion of k under||. The mapping a→(a, a, a,...) is an isomorphismof k in ¯k and we identify this isomorphic image with k itself.We extend to ¯k the valuation|| in k, in the following manner. The realnumber field ¯Γ is complete under the valuation induced by the uniqueorder on it. So, if a∈ ¯k, then a=(a 1 , a 2 ,...), a Cauchy sequence ofelements of k. Put|a|= limn→∞|a n |That this is a valuation follows from the properties of limits in ¯Γ.Also, the extend valuation is archimedian or non-archimedian, accordingas the valuation in k is archimedian or non-archimedian.For instance, if|| on k is non-archimedian anda=(a 1 , a 2 ,...)b=(b 1 , b 2 ,...)are two elements of ¯k, thenanda+b=(a 1 + b 1 , a 2 ,+b 2 ,...)|a+b|−Max(|a|,|b|)= limn→∞(|a n + b n |−Max(|a n |,|b n |))which is certainly≤0,¯k is thus a complete valuated field.It may also be seen that the elements of k are dense in ¯k in the topol- 211ogy induced by the metric.Let c 1 + c 2 + c 3 +··· be a series in ¯k. We denote by S n the partialsumS n = c 1 + c 2 +···+c n .We say that c 1 + c 2 +···+c n +··· is convergent if and only if thesequence of partial sums S 1 , S 2 ,...,S n ,... converges. This means that,for everyε>0, there exists an integer n=n(ε) such that|S m ′− S m |=|c m+1 +···+c m ′| n.

186 8. Valuated fieldsObviously c 1 , c 2 ,... is a null sequence in ¯k.In case the valuation is non-archimedian, we have the followingproperty:-1) The series c 1 + c 2 +···+c n +··· is convergent if and only ifc 1 , c 2 ,... is a null sequence.Proof. We have only to prove the sufficiency of the condition. Supposethat c n → 0; then, for large n and m,|S n − S m |=|c m+1 + c m+2 +···+c n |≤ max(|c m+1 |,...,|c n |)and so tends to zero. This proves the contention.□212Note that this theorem is false, in case the valuation is archimedian.Let k be a field and|| a valuation on it. a→|a| is a homomorphismof k ∗ into multiplicative group of positive real numbers. Let G(k) denotethis homomorphic image. This is a group which we call the value groupof k for the valuation. If ¯k is the completion of k by the valuation in k,then G(¯k) is value group of ¯k.Suppose|| is an archimedian valuation. Then k has characteristiczero and containsΓ, the rational number field as a subfield. OnΓ,|| isthe ordinary absolute value. Since ¯k contains ¯Γ, the field of real numbers,it follows that1) G(¯k) is the multiplicative group of all positive real numbers.Also, because of the definition of the extended valuation, it followsthat G(k) is dense in the group G(¯k).We shall now assume that|| is a non-archimedian valuation. We considerthe exponential valuation. Then G(k) is a subgroup of the additivegroup of all real numbers. We shall now prove2) G(¯k)=G(k).Proof. For, let 0a∈ ¯k. Then a=(a 1 ,...) is a Cauchy sequence,not a null sequence in k. By definition,|a|=limn|a n |

4. Complete fields 187Now a n = a n − a+a so that|a n |≥min(|a n − a|,|a|).But, for n large,|a n − a|>|a| so that|a n |=|a| and our contention isestablished.□G(k) being an additive subgroup of the real number field is eitherdense or discrete. The valuation is then called dense or discrete accordingly.In the second case, there existsπin k with smallest pos- 213itive value|π|·|π| is then the generator of the infinite cyclic groupG(k)·π is called a uniformising parameter. It is clear thatπis notunique. For, if u∈k with|u|=0, then|uπ|=|π| and uπ is also auniformising parameter.Consider in k the set O of elements a with|a| ≥ o.O is then anintegrity domain. For,|a|≥0,|b|≥0⇒|a+b|≥Min(|a|,|b|)≥0.Also|ab| = |a|+|b| ≥ 0. We call O the ring of integers of thevaluation. Consider the set Y of elements a∈k with|a|>0. ThenY is a subset of O and is a maximal ideal in O. For, if a∈r andb∈Y , then|ab|=|a|+|b|>0. Also, if a∈r but not in Y , then|a|=0 and|a −1 |=0 so that if U is an ideal in O containing Y , thenU = Y or U = r. We call Y , the prime divisor of the valuation.Since Y is maximal, r/Y is a field. We callπ/Y the residue classfield.Exactly the same notions can be defined for ¯k. We denote by O¯thering of integers of the valuation so that O ¯ is the set of a∈ ¯k with|a|≥0. Y¯is the maximal ideal in O, ¯ hence the set of a∈ ¯k with|a|>0. Also O/ ¯ Y ¯ is the residue class field. ClearlyY = ¯ Y ∩ rWe now have

188 8. Valuated fields2) Every a∈k(¯k) has the property; a∈r(¯r) or a −1 ∈ Y ( ¯ Y ).This is evident since if aO( O),|a|0 and hence 214a −1 ∈ Y ( Y ¯ ).3) O( ¯ O) is integrally closed in k(¯k).Proof. We should prove that everyαin k(¯k) which is a root of apolynomial of the type x n + a 1 x n−1 +···+a n , a 1 ,...,a n ∈ O( O), ¯ isalready in O( O). ¯ For, supposeα n + a 1 α n−1 +···+a n = oandαO( ¯ O). Thenα −1 ∈ Y ( ¯ Y ). Therefore1=−(a 1 α −1 + a 2 α −2 +···+a n α −n )and soo=|1|≥min(|a 1 α −1 |,...|a n α −n )>owhich is absurd□4) Every element in ¯ O is the limit of a sequence of elements in O andconversely.Proof. Let a=(a 1 ,...,a n ,...) be in ¯ O, a 1 ,...,a n ,... in k. Then, aswe saw earlier, for sufficiently large n|a n |=|a|.But|a|≥0 so that|a n |≥0. Thus, for sufficiently large n, all a ′ n s arein O. Converse is trivial.We now prove5) There is a natural isomorphism of O/Y on ¯ O/ ¯ Y .□

4. Complete fields 189215Proof. The elements of O/Y are residue classes a+Y , a∈r. Wenow make correspond to a+Y the residue class a+ Y ¯ . Then a+ Y ¯is an element of O/ ¯ Y ¯. The mapping a+Y → a+ Y¯is a homomorphism(non-trivial) of O/Y into ¯℧/ Y ¯. Since both are fields, thisis an isomorphism into. In order to see that it is an isomorphism ofO/Y onto O/ ¯ Y ¯ , let ā+ Y ¯ be any residue class in O/ ¯ Y ¯ . By (4)therefore, given any N> 0, there exists a∈℧with|ā−a|>N> 0or ā−a∈ Y ¯. If we then take a+Y, then ā+ Y¯is the image ofa+Y . This proves that the mapping is an isomorphism onto. □(5) enables us to choose, in k itself, a set of representatives of ¯℧/ ¯ Y ,the residue class field. We shall denote this set by R and assume thatit contains the zero element of O. Also it has to be observed that, ingeneral, R is not a field. It is not even an additive group.We now assume that|| is a discrete valuation. Letπin O be a uniformisingparameter. Then clearly¯ Y = (π)is a principal ideal. Let a∈ ¯k. Then|a|/|π| is a rational integer, sinceG(k) is an infinite cyclic group. Put|a|/|π|=t. Then|aπ −t |=0. Ifwe call elements u in ¯k with|u|=0, units, thenwhere u is a unit.a=π t uLet R be the set defined above and a∈ ¯℧. Thena≡a 0 ( mod ¯ Y )with a 0 ∈ R. This means that (a−a 0 )π −1 is an integer (in ¯Ω). 216(a−a 0 )π −1 ≡ a 1 (mod ¯ Y ),

190 8. Valuated fieldswhere a 1 ∈ R. Thena≡a 0 + a 1 π(mod ¯ Y 2 )In this way, one proves by induction thata≡a 0 + a 1 π+···+a m π m (mod) ¯ Y m+1 )where a 0 , a 1 ,...,a m ∈ R. Put b m = a 0 + a 1 π+···+a m π m .Then a≡b m (modY ¯m+1) which means thatConsider the series|a−b m |≥m+1b 0 + (b 1 − b 0 )+(b 2 − b 1 )+···Then, since b m+1 − b m = a m+1 π m+1 , we see that|b m+1 − b m | increasesindefinitely. Hence the above series converges. Also, since its elementsare integers,is an element of ¯ O.b=b 0 + (b 1 − b 0 )+(b 2 − b 1 )+···Since b 0 + (b 1 − b 0 )+···+(b m − b m−1 )=b m , it follows thatb=limmb m .Thus we havea= limm→∞ b m= a 0 + a 1 π+a 2 π 2 +···217By the very method of construction this expression for a is unique,once we have chosen R andπ.If a∈ ¯k, aπ t ∈ O ¯ for some rational integer t. Hence we have

4. Complete fields 1916) Every element a in ¯k has the unique expressiona=∞∑a n π nn=−ta n ∈ R andπin ¯ O is a generator of Y . t is a rational integer.If t>0, we shall callh(a)=∑−1n=−ta n π nthe principal part of a. If t≤0 we put h(a)=0. Clearly, h(a) is anelement in k. Also a−h(a)∈ O. ¯We now study the two important examples of the rational numberfield and the rational function field of one variable.LetΓbe the field of rational numbers. We shall denote by|| ∞ theordinary absolute value and by|| p the p-adic value for p, a prime. Ifa is an integer, a= p λ n 1 , (p, n 1 )=1.We put|a| p =λ log pand|a| ∞ = absolute value of a.It is then clear that each of the non-archimedian valuations ofΓisdiscrete. Let us denote byΓ ∞ the completion ofΓby the archimedianvaluation and byΓ p the completion ofΓby the p-adic valuation.Γ ∞is clearly the real number field.If O p denotes the set of integers ofΓ p and Y the prime ideal of thevaluation thenY = (p)A set of representatives of O p mod Y is given by the integers 0, 1, 2, 218..., p−1 as can be easily seen. Hence, by (6),

192 8. Valuated fields7) Every a∈Γ p is expressed uniquely in the formwhere a i = 0, 1, 2,..., p−1.a=∞∑a n p nn=−tThe elements ofΓ p are called the p-adic numbers of Hensel.As before, we denote, by h p (a), the principal part of a at p. Clearlya−h p (a) is a p-adic integer.Let a be a rational number, a ∈ ℧ p ⇔ |a| p ≥ 0, that is a = b c ,(b, c)=1 and c is prime to p. Since only finitely many primes divideb and c, it follows that a∈℧ p for almost all p, that is except for afinite number of p. Hence h p (a)=0 for all except a finite number ofprimes. Hence for any a∑h p (a)phas a meaning. Also, a−h p (a) is a rational number whose denominatoris prime to p. Hence a− ∑ h p (a) is a rational number whosepdenominator is prime to every rational integer. Hence8) For every rational number a∑a− h p (a)≡0(mod1).p219This is the so-called partial fraction decomposition of a rationalnumber.Let now k be an algebraically closed field and K = k(x) the fieldof rational functions of one variable x. All the valuations are nonarchimedian.Every irreducible polynomial of k[x] is linear and ofthe form x−a. With every a∈kthere is the valuation|| a associated,which is defined by| f (x)| a =λ,

4. Complete fields 193where f (x)∈k[x], f (x)=(x−a) λ ϕ(x),ϕ(a)0. If we denote by|| ∞ the valuation by degree of f (x), then, for f (x)∈k[x],| f (x)| ∞ =− deg f (x)Let K a and K ∞ denote the completions, respectively, of K at|| a and|| ∞ . If O a and O ∞ are the set of integers of K a and K ∞ , Y a and Y ∞the respective prime divisors, thenY a ={x−a}, Y ∞ ={ 1 x }.It is clear, then, that℧ a /Y a and℧ ∞ /Y ∞ are both isomorphic to k andsince K contains k, we may take k itself as a set of representatives ofthe residue class field. Any element f in K a is uniquely of the formf=a n ∈ k. Similarly, ifϕ∈K ∞ ,ϕ=∞∑a n (x−a) n ,n=−t∞∑b n x −n .n=−tAs before, if we denote by h a ( f ) and h ∞ ( f ) the principal parts off∈ K for the two valuations, then∑h a ( f )+h ∞ ( f )ahas a meaning since h a ( f )=0 for all but a finite number of aIf we defineϕ∈K to be regular ar a(∞) ifϕ∈℧ a (℧ ∞ ), then for 220f∈ K, ∑f− h a ( f )awhere a may be infinity also, is regular at all, a∈kand also for thevaluation|| ∞ . Such an element, clearly, is a constant. Hence

194 8. Valuated fields9) If f∈ K then∑f− h a ( f )= constantaConversely, it is easy to see that there exists, up to an additive constant,only one f∈ K which is regular for all a∈kexcept a 1 ,...,a n(one of which may be∞also) and with prescribed principal parts atthese a i . 9) gives the partial fraction decomposition of the rationalfunction f .5 Extension of the valuation of a completenon-archimedian valuated field221We shall study the following problem. Suppose k is compute under avaluation||. Can this valuation be extended, and if so in how manyways, to a finite algebraic extension K of k ?We prove, firstLemma 1. Let k be a field complete under a valuation||, and K a finitealgebraic extension of k. Letω 1 ,...,ω n be a basis of K/k. Let|| havean extension to K andn∑α ν = a iv ω i , a iv ∈ k,i=1v=1, 2, 3,... be a Cauchy sequence in K. Then a iv i=1, 2,...,nare Cauchy sequence in K.Proof. We consider the Cauchy sequence{α v },m∑α v = a iv ω i , a iv ∈, ki=11≤m≤n and we shall prove that the{a iv } are Cauchy sequences in k.We use induction on m. Clearly, if m=1,α v = a iv ω 1and{α v } is a Cauchy sequence in K if and only if{a iv } is a sequence ink. □

5. Extension of the valuation of a complete. . . 195Suppose we have proved our statement for m−1≥1, instead of m.Writem−1 ∑α v = a iv ω i + a mv ω m .i=1If{a mv } is a Cauchy in k, then{α v − a mv ω m } is a Cauchy sequencesin K and induction hypothesis works. Let us assume that a mv is not aCauchy sequence in k. This means that there exists aλ>0 and for everyv, an integerµ v such thatµ ν > vandK.|a m µ v − a mv |>λ.Consider now the sequence{β v } in K withβ v = α µ v−α va mµv − a mv.Because of the above property, we see that{β v } is a null sequence inNowm−1 ∑( )ai µ v − a ivβ v −ω m = ω ia m µ v − a mvi=1and{β v −ω m } is a Cauchy sequence in K. Induction hypothesis now 222works and so, if ( )aiµv − a ivlim = b i ,v→∞ a mµv − a mvthen∑−ω m = b i ω i , b i ∈ k.i=1This is impossible becauseω 1 ,...,ω m are linearly independent overk. Therefore{a mv } is a Cauchy sequence and our Lemma is therebyproved.We shall now prove the following theorem concerning extension of 223valuation.

196 8. Valuated fieldsTheorem 4. If the valuation|| of a complete field k can be extended to afinite extension K, then this extension is unique and K is complete underthe extended valuation.Proof. That K is complete under the extended valuation is easy to see.For, if{α v } is a Cauchy sequence in K and∑α v = a iv ω i , a iv ∈ k.,by the lemma, the{a iv }’s are Cauchy sequences. So, ifilim a iv= b i ∈ k,v→∞thenwhich is again in K.lim α γ=v→∞∑iω 1 limv→∞a iv =∑b i ω ii□We shall now prove that the extended valuation is unique.From the lemma, it follows that if{α v } is a null sequence in K andα v = ∑ a iv ω i , a iv ∈ k, then the a iv ’s are null sequences in k.iIn particular, ifα∈Kand|α|

5. Extension of the valuation of a complete. . . 197In a similar manner, if|α|>1, then|Nα|>1. Thus we get|Nα|=1⇒|α|=1.Let nowβbe in K and writeβ= αnwhere n=(K : k)NαThen Nβ= (Nα)n(Nα) n . Thus|Nβ|=1.By the above, it means that|β|=1in the valuation. Hence|α|= n√ |Nα|showing that the value ofαin the extended valuation is unique fixed. 224Our theorem is thus completely proved.In order to prove that an extension of the valuation is possible, weshall consider the case where k is complete under a discretenon-archimedian valuation. Let O be the ring of integers Y , the primedivisor of the valuation and O/Y , the residue class field. We shall nowprove the celebrated lemma due HenselLemma 2. Let f (x) be a polynomial of degree m in O[x], g 0 (x), a monicpolynomial of degree r≥1 and h 0 (x), a polynomial of degree≤m−rboth with coefficients in O such that1) f (x)≡g o (x)h o (x) (modY )2) g o (x) and h o (x) are coprime mod YThen there exists polynomials g(x) and h(x) in O[x] such that⎫g(x)≡g o (x) ⎪⎬(mod Y ),h(x)≡h o (x)⎪⎭g(x) has the same degree as g o (x) and f (x)=g(x)· h(x).Proof. We shall now construct two sequencesg o (x), g 1 , (x),... and h o (x), h 1 , (x),... satisfyingof polynomialsg n (x)≡g n−1 (x)( mod Y n )

198 8. Valuated fieldsh n (x)≡h n−1 (x)( mod Y n )f n (x)≡g n (x)h n (x)( mod Y n+1 )g n (x) is a monic polynomial of degree r and h n (x) of degree≤ m−r. Allthe polynomials have coefficients in O. □225The polynomials are constructed inductively. For n=0, g o (x) andh o (x) are already given and satisfy the conditions. Assume now thatg o (x),...,g n−1 (x) and h o (x),...,h n−1 (x) have been constructed so assatisfy the requisite conditions.Since g n−1 (x) ≡ g o (x)(mod Y ) and h n−1 (x) ≡ h o (x)(mod Y ) andg 0 (x) and h o (x) are coprime mod Y , there exists, for any polynomialf n (x) in O[x], two polynomials L(x) and M(x) withf n (x)≡L(x)g n−1 (x)+ M(x)h n−1 (x) (mod Y ).L(x) and M(x) are clearly not uniquely determined. We can replace L(x)by L(x)+λ(x)h n−1 (x) and M(x) by M(x)+λ(x)g n−1 (x).Letπbe a generator of the principal ideal Y . By induction hypothesis,f n (x)=π −n ( f (x)−g n−1 (x)h n−1 (x))is an integral polynomial, so in O[x]. Since g n−1 (x) has degree r and ismonic and h n−1 (x) degree≤ m−r, it is possible to choose M(x) and L(x)so that M(x) has degree

5. Extension of the valuation of a complete. . . 199g(x)=g o (x)+(g 1 (x)−g o (x))+···+(g n (x)−g n−1 (x))+···h(x)=h o (x)+(h 1 (x)−h o (x))+···+(h n (x)−h n−1 (x))+···Since g n (x)−g n−1 (x)=0(modπ n ), it follows that the correspondingcoefficients of the sequence of polynomials g 0 (x), g 1 (x),... formCauchy sequences in O. Since k is complete andg(x)= limn→∞g n (x),it follows that g(x)∈℧[x]. It is monic and is of degree r. In a similarway, b(x)∈℧[x] and has degree≤m−r.Also since f (x)−g n (x)h n (x)≡0(mody n+1 ), it follows that the coefficientsof ( f (x)−g n (x)h n (x)) form null sequences. Hencef (x)=limng n (x)h n (x)=h n (x)=g(x)h(x)and our lemma is proved.We now deduce the following important .Lemma 3. Let f (x)= x n +a 1 x n−1 +···+a n be an irreducible polynomialk[x] , k satisfying hypothesis of lemma 2. Then f (x)∈O[x] if and onlyif a n ∈ O.Proof. It is clearly enough to prove the sufficiency of the condition. Leta n ∈℧, and if a 1 ,...,a n−1 (some or all of them) are not in℧, then thereis a smallest powerπ a , a>0, ofπsuch thatπ a f (x)=b o x n + b 1 x n−1 +···+b nis a primitive polynomial in O[x]. Also, now b n ≡ 0( mod Y ). and 227at least one of b 0 ,...,b n−1 is not divisible by Y . Let b r be the firstcoefficient from the right not divisible by Y . Thenπ a ≡ (b 0 x r +···b r )x n−r (mod Y )Since b r 0( mod Y ), Hensel’s lemma can be applied and we seethatπ a f (x) is reducible in O[x]. Thus f (x) is reducible in k[x] whichcontradicts the hypothesis. The lemma is therefore established. □

200 8. Valuated fieldsWe are now ready to prove the important theorem concerning extensionof discrete non-archimedian valuations, namelyTheorem 5. Let k be complete under a discrete non-archimedian valuation||and K a finite algebraic extension over k. Then|| can be extendeduniquely to K and then for anyαin K,|α|=1(K : k) |N K/kα|.Proof. Because of Theorem 4, it is enough to prove that the functiondefined on K by1|α|=(K : k) |N K/kα|is a valuation function. Clearly ,|0|=∞;|α| is a real number forα0.Also,We shall now prove that|αβ|=|α|+|β|.|α+β|≥min(|α|,|β|).228 ∣ Ifαorβis zero, then the above is trivial. So letα0,β0. Since∣∣∣∣ α∣ ∣∣∣∣ β∣ or βα∣ is≥ 0, it is enough to prove that if|λ|≥0,|1+λ|≥0. □Let f (x)= x m + a 1 x m−1 +···+a m be the minimum polynomial ofλin K over k, ThenNλ=((−1) m a m ) (K:k(λ)) .Also, N(1+λ) = (−1) n (1±a 1 ±···+a m ) (K:k(λ)) . If|Nλ ≥ 0|,then|a m | ≥ 0 which , by lemma 3, means that|a 1 |,...,|a m |. Hence|N(1+λ)|≥0. Our theorem is proved.Incidentally it shows that the extended valuation is discrete also.

6. Fields complete under archimedian valuations 2016 Fields complete under archimedian valuationsSuppose k is complete under an archimedian valuation. Then k has characteristiczero and contains, as a subfield, the completion ¯Γ of the rationalnumber field. barΓ(i) is, then, the complex number field. Everycomplex number is to the form a+ib, a, b∈ ¯Γ. On ¯Γ, we have theordinary absolute value. Define in ¯Γ(i) the function|z|=(a 2 + b 2 ) 1 2where z=a+ib. It is, then, easy to verify that|| is a valuation on ¯Γ(i)which extends the valuation in ¯Γ. Also, by theorem 4, this is the onlyextension of the ordinary absolute value. We consider the case k⊃ ¯Γ(i)and prove the theorem of A. Ostrowski.Theorem 6. Let k⊃ ¯Γ(i) be the complex number field and let k be afield with archimedian valuation and containing ¯Γ(i). If the valuation ink is an extension of the valuation in ¯Γ(i), then k= ¯Γ(i).Proof. If k ¯Γ(i), let a∈kbut not in ¯Γ(i). Denote by|| the valuation in 229k. Consider|a−z| for all k= ¯Γ(i). Since|a−z|1≥0, we haveρ=g· 12· b|a−z|≤0.There exists, therefore, a sequences z 1 ,...,z n ,.. of complex numberssuch thatlimn→∞ |a−z n|=ρ.But z n = z n − a+a and so|z n |≤|z n − a|+|a| which shows that the|z n |, for large n, are bounded. We may therefore, choose a subsequencez i1 , z i2 ,... converging to a limit point z o such thatρ= limn→∞|a−z in |=|a−z o |We have thus proved the existence of a z 0 in ¯Γ(i) such that b=a−z 0has|b|=ρ. Since, by assumption, a ¯Γ(i) we have|b|=ρ>0.□

202 8. Valuated fieldsρ, by definition, being g.l.b., it follows that|b−z|≥ρfor z∈ ¯Γ(i).Consider the set of complex numbers z with|z|0 be anarbitrary rational integer andε, a primitive nth root of unity. Thenb n − z n = (b−z)(b−εz)...(b−ε n−1 z).230ThereforeHence|b−z|ρ n−1 ≤|b−z||b−εz|...|b−ε n−1 z|=|b n − z n |≤|b| n +|z| n =ρ n (1+( |z|ρ )n ).)|b−z|≤ρ(1+ |z|nρ n .But|z|

6. Fields complete under archimedian valuations 203which shows that all complex numbers are bounded in absolute value.This is a contradiction. Hence our assumption that a ¯Γ(i) is false .The theorem is thereby proved,Before proving theorem 7 which gives a complete characterizationof all complete fields with archimedian valuation, we shall prove a coupleof lemmas.Lemma 4. Let k be complete under an archimedian valuation|| andλ 231in k such that x 2 +λ is irreducible in k[x]. Then|1+λ|≥1.Proof. If possible, let|1+λ| < 1. We construct, by recurrence, thesequence c 0 , c 1 , c 2 ,..., in k, defined as follows: -c 0 = 1c n+1 =−2− 1+λc nn=0, 1, 2,...It, then, follows that|c n |≥1. For, if we have proved it upto c n−1 ,then|c n |≥2− |1+λ| ≥ 1.|c n−1 |Thus c n does not vanish for any n. Also,|c n+1 − c n |= |1+λ||c n− c n−1 ||c n ||c n−1 |≤ρ|c n − c n−1 |whereρ=|1+λ|

204 8. Valuated fieldsLemma 5. If k is complete under an archimedian valuation,||, then thisvaluation can be extended to k(i).232Proof. If i∈k, there is nothing to prove. Let ik. Then every elementof k(i) is of the form a+ib, a, b∈k.The norm from k(i) to k ofα=a+ib isNα=a 2 + b 2 .By theorem 4, therefore, it is enough to prove that|α|=|(a 2 + b 2 ) 2 1 |is a valuation on k(i). By puttingλ= b2in the lemma 4, we see thata2 |a 2 + b 2 |≥a 2 . Therefore|(1+a) 2 + b 2 |≤1+|a 2 + b 2 |+2|a|≤ 1+|a 2 + b 2 |+2 √ |a 2 + b 2 |= (1+ √ |a 2 + b 2 |) 2 .This shows thatand our lemma is proved.|1+α|≤1+|α|□We now obtain a complete characterization of complete archimedianfields, namely,Theorem 7. The only fields complete under an archimedian valuationare the real and complex numbers fields.Proof. k has characteristic zero and since it is complete, it contains thefield ¯Γ of real numbers. If k contains ¯Γ properly, then we assert that kcontains i. For, k(i), by lemma 5, is complete and k(i) contains ¯Γ(i) and,by theorem 6,k(i)= ¯Γ(i).□

7. Extension of valuation of an incomplete field 205Therefore¯Γ(i)=k(i)⊃k⊃ ¯Γ.But (¯Γ(i) : ¯Γ)=2so that k=k(i)= ¯Γ(i).We have thus found all complete fields with archimedian valuation.7 Extension of valuation of an incomplete fieldSuppose k is a complete field under a valuation|| and letΩbe its alge- 233braic closure. Then|| can be extended toΩby the prescription|α|=|Nα| 1 n ,where Norm is takes form k(α) over k and n=(k(α) : k). It is clear thatit defines a valuation function. For, of K is a subfields ofΩand K/k isfinite and K containsα then, by properties of norms,|α|=|N K/k α| 1 m ,where m=(K : k). So, ifαandβare inΩ, we may take for K a fieldcontainingαandβand with (K : k) finite.Furthermore, defined as such, the valuation onΩis dense because,if|α|>1, then|α| 1/n has value as near 1 as one wishes, by increasing nsufficiently. Also, for every n,α 1/n is inΩ.Also, letσbe an automorphism ofΩ/k, andαinΩ. Then, by definitionof norm,Nα=N(σα)so that|α|=|σα|. Thus all conjugates of an element have the samevalue.We shall now study how one can extend a valuation of an incompletefield to an algebraic extension.Let k be a field and K a finite algebraic extension of it. Let|| be avaluation of k and ¯k the completion of k under this valuation. Let k ¯k. 234

206 8. Valuated fields KkSuppose it is possible to extend|| to K. Let ¯K be the completion ofK under this extended valuation. Since ¯K⊃K⊃ k, it follows that ¯Kcontains K and ¯k and therefore the composite K¯k. Thus¯K⊃ K¯k.On the other hand, K ¯k/¯k is a finite extension, since K/k is finite.Since ¯k is complete, K¯k is complete also. K¯k contains K and hence itscompletion ¯K under this extended valuation. Thus¯K= K¯k.Thus if the valuation can be extended, then the completion of K bythis extended valuation is a composite extension of K and ¯k.Suppose now thatΩis an algebraic closure of ¯k.Ω, then, containsan algebraic closure of k. We have seen above that the given valuationof k can be extended toΩ. Letσbe an isomorphism of K/k intoΩ.The valuation in ¯k can be extended toσK·¯k which is a subfield ofΩ.Therefore, there is a valuation onσK. Define now, forαin K,K ¯k¯k|α| o =|σα|235where|| is the extension of|| on k toσK·¯k, which extension is unique.It is now trivial to see that|| o is a valuation on K and extends|| on k.Hence every isomorphism of K intoΩwhich is trivial on k, givesrise to a valuation of K.We now inverstigate when two isomorphisms give rise to the samevaluation on K. Letσandτbe two isomorphisms of K/k intoΩgivingthe same valuation on K.σK andτK are subfields ofΩand they have

7. Extension of valuation of an incomplete field 207the same valuation. Thusµ=στ −1 is an isomorphism ofτK ontoσKwhich preserves the valuation onτK. Nowµis identity on k and so on ¯k.SinceσK·¯k is the completion ofσK, it follows thatµis an isomorphismof the composite extensionsσK¯k andτK¯k. Hence, ifσandτgive riseto the same valuation on K, the corresponding composite extensions areequivalent.Suppose now thatσandτare isomorphisms of K intoΩsuch thatthe composite extensionsσK¯k andτK¯k are equivalent. There exists thena mappingµofτK¯k onσK¯k which is identity on ¯k and such thatµτ=σσ induces a valuation|| 1 on K such that|α| 1 =|σα| andτinduces avaluation|| 2 on K such that|α| 2 =|τα|. Butµis such thatµτα=σα orσα andτα are conjugates over ¯k inΩ. Hence|σα|=|τα|or|| 1 =|| 2 which shows thatσ,τgive the same valuation on K.We have hence theTheorem 8. A valuation|| of k can be extended to a finite extension Kof k only in a finite number of ways. The number of these extensions of 236|| to K stand in a (1, 1) correspondence with the classes of compositeextensions of K and ¯kFrom what we have already seen, the number of distinct compositeextensions of K and ¯k is at most (K : k).We apply these to the case where k=Γ, rational number field andK/Γ finite so that K is an algebraic number filed. From theorem 2,it therefore follows that K has at most (K : Γ) distinct archimedianvaluations and that all the non-archimedian valuations of K, which arecountable in number, are discrete.In a similar manner, if K is an algebraic function field of one variableover a constant filed k, then all the valuations of K are non-archimedianand discrete. This can be seen from the fact that if x∈ K is transcendentalover k, then K/k(x) is algebraic and one has only to apply theorems3 and 8.

AppendixAbelian groups1 Decomposition theoremAll the groups that we deal with here are abelian. Before proving themain decomposition theorem for finite abelian groups we shall provesome lemmas.Lemma 1. If a, b are elements in G and have orders m and n respectivelyand (m, n)=1, then ab has order mn.Proof. Clearly if t is the order of ab, t|mn, since237(ab) mn = (a m ) n (b n ) m = e,e being unit element of G. Also a t = b −t . Thuse=a tm = b −tmso that n/t. Similarly m|t. Hence t=mn.□Lemma 2. Let p be a prime number dividing the order n of the groupG. Then there is, in G, an element of order p.Proof. We use induction on n. Let a be an element in G of order m. Ifp|m, then a m/p has order p and we are through. Suppose p∤m. Let Hbe the cyclic group generated by a· G/H has then order n/m which isdivisible by p. Since n |n, induction hypothesis applies, so that there isma coset Hb of order p. If b has order t then b t = e and so (Hb) t = Hwhich means that p|t and so b t/p has order p. □209

210 8. AppendixLemma 3. Let G be a finite group andλthe maximum of the orders ofelements of G. Thena λ = e238for all a∈G.Proof. Let b be an element of orderλ. Let a be any element in G andletµbe its order. To prove the lemma, it is enough to prove thatµ|λ. Ifnot, there is a prime p which dividesµto a higher power than it doesλ. Let p r be the highest power of p dividingµand p s the highest powerdividingλ. Then r>s. We will see that this leads to a contradiction. □Since µ p r and pr are coprime, a µ p rhas the order p r . Similarly b ps hasorder λ p s . By lemma 1, c=a µ p r· b ps239has order p r λ.ps>λ, which contradicts the definition ofλ. Henceµ|λ.Lemmas 2 and 3 show thatλ|n where n is the order of the group andthatλand n have the same prime factors.λ is called the exponent of thefinite group G.A set a 1 ,...,a n of elements of a finite group G are said to be independentifa x 11 ···ax nn = eimplies a x ii= e, i=1,...,n.If G is a finite group and is a direct product of cyclic groups G 1 ,..., G n and if a i , i=1,...,n is a generator of G i , then a 1 ,...,a n areindependent elements of G. They are said to form a base of G.We shall now proveTheorem 1. Let G be a finite group of order n. Then G is the directproduct of cyclic groups G 1 ,...,G l of ordersλ 1 ,...,λ l such thatλ i |λ i−1 , i=2,...,l,λ l > 1.Proof. We prove the theorem by induction on the order n of the groupG. Let us therefore assume theorem proved for groups of order< n. Let

1. Decomposition theorem 211G have order n. Supposeλ 1 is the exponent of G. Ifλ 1 = n, then Gis cyclic and there and there is nothing to prove. Let thereforeλ 1 < n.There is an element a 1 of orderλ 1 . Let G 1 be the cyclic group generatedby a 1 . G/G 1 has order n λ 1< n. Hence induction hypothesis works onG/G 1 .G/G 1 is thus the direct product of cyclic groups W 2 ,...,W l of orderλ 2 ,...,λ l respectively andλ l |λ l−1 |...|λ 2 . Let H i be a generator of W i .Then H i = G l b i for some b i in the coset H i . Let the element b i in G haveorder t i . Then t i |λ 1 by lemma 3. ButH t ii= b t ii Gt i1 = G 1which proves thatλ i |λ 1 . Thusλ l |λ l−1 |...|λ 1 .Let now b λ ii= a x i1 . Put x i= y i· z i where (y i ,λ 1 )=1 and all the primefactors of z i divideλ 1 . Choose u i prime toλ 1 such thatu i y i ≡ 1(modλ 1 ).□Then b u iλ ii= a z i1 . Sinceλ 1 is the exponent of G, 240e=b u iλ 1i= (a z 11 )λ 1/λ i.Since a 1 has orderλ 1 this means thatλ i |z i , i=2, 3,...,l.Put nowa i = b u iia −z i/λ i1.Since u i is prime toλ 1 and so toλ i , the coset F i = G l a i is also agenerator of W i . Thus G 1 a 2 ,...,G 1 a l is a base of G/G 1 .Let a i have order f i . Thene=a f ii= b u i f iia − f iz i /λ ii.This means thatb u i f ii= a f iz i /λ ii.Therefore by definition of b i ,λ i |u i f i . But (u i λ i )=1. Henceλ i | f i .

212 8. Appendix241242On the other handa λ ii= b u iλ iia −z ii= e.Hence f i =λ i . We have thus elements a 1 ,...,a l in G which haveordersλ 1 ,...,λ l satisfyingλ l |λ l−1 |...|λ 1 .We maintain that a 1 ,...,a 1 are independent elements of G. For, ifa v 11 ...av ll= e,then a v 22 ·····av ll= a −v 11which means that F v 22 ...Fv ll= G 1 . But sinceF 2 ,..., F l are independent,λ i |v i , i=2,...,l. But this will mean thata v 11 = e orλ 1|v 1 .Sinceλ 1 ...λ l = n, if follows that a 1 ,...,a l form a base of G andthe theorem is proved.Let G be a group of group of order n and let G be direct product ofcyclic groups G 1 , G 2 ,...,G l of ordersλ 1 ,...,λ l . We now proveLemma 4. Letµbe a divisor of n. The number N(µ) of elements a∈Gwitha µ = eis given byN(µ)=l∏(µ,λ i ).i=1Proof. Let a 1 ,...,a l be a base of G so that a i is of orderλ i . Any a∈Ghas the forma=a x 11 ···ax ll .If a µ = e, then e=a x 1µ. Since a 1 ,...,a l is a base, this means that1···a x lµlx i µ≡0(modλ i ), i=1,λ, l. Hence x i has precisely (µ,λ i ) possibilitiesand our lemma is proved.We can now prove the □Theorem 2. If G is the direct product of cyclic groups G 1 ,... G l , ofordersλ 1 ,...,λ l respectively withλ l |λ l−1 |···|λ 1 and G is also the directproduct of cyclic groups H 1 ,..., H m of ordersµ 1 ,...,µ m respectivelywithµ m |µ m−1 |···|µ 1 , then m=1andλ i =µ i , i=1,...,l.

2. Characters and duality 213Proof. Without loss in generality let l≥m. Let a 1 ,...,a l be a base of Gin the decomposition G 1 × G 2 ×···×G 1 . Since the number of elementsa with a µ = e is independent of the decompositionl∏N(µ)= (µ,λ i )=1=1m∏(µ,µ j )j=1Put nowµ=λ 1 . Then N(µ)=λ 1 1 . But since (µ,µ j)≤µ, it followsthatλ l l ≤λm l, so that l≤m. This provesl=m.Also it follows that each factor (λ 1 ,µ j )=λ l orλ l |µ l . Inverting theroles ofλandµwe getλ l =µ l .Suppose now it is proved thatλ q+1 =µ q+1 ,...,λ 1 =µ 1 . Then byputtingµ=λ q , we have□N(µ)=λ q q.λ q+1···λ 1 =q∏(λ q ,µ j )λ q+1···λ l .j=1By the same reasoning as before, it follows thatλ q =µ q and we are,therefore, through.For this reason, the integersλ 1 ,...,λ l are called the canonical invariantsof G. From theorems 1 and 2 we have the Corollary Two finitegroups G and G ′ are isomorphic if and only if they have the same canonicalinvariants.2 Characters and dualityLet G be a group, not necessarily abelian and Z a cyclic group. A homomorphismχofG into Z is called a character of G. Thus243χ(a)χ(b)=χ(ab).

214 8. AppendixIf we denote the unit element of G by e and that of Z by 1, thenχ(e)=1.The characterχ o defined byχ o (a)=1 for all a∈G is called thePrincipal character.Ifχ 1 andχ 2 are two characters, we define their productχ=χ 1 χ 2 byand the inverse ofχ 1 byχ(a)=χ 1 (a)χ 2 (a)χ −11 (a)=(χ 1(a)) −1 .Under this definition, the characters form a multiplicative abeliangroup G ∗ called the character group of G.Sinceχis a homomorphism, denote by G χ the kernel of the homomorphismχof G into Z. Then G/G χ is abelian. Denote by H thesubgroup of G given by⋂H= G χ , χ∈G ∗ .χ244Then clearly G/H is abelian.We call Z an admissible group for G if H consists only of the identityelement. This means first that G is abelian and furthermore that givenany two elements a, b in G there exists a characterχ of G such that, ifab,χ(a)χ(b).Ifχis a character of G, thenχcan be considered as a character ofG/G o whereχis trivial on G o , by definingχ(G o a)=χ(a), G o a being acoset of G modulo G o . In particular, the elements of G ∗ can be consideredas characters of G/H. Moreover Z is now an admissible group forG/H.We now prove theTheorem 3. If G is a finite abelian group and Z is an admissible groupfor G, then G ∗ is finite and G is isomorphic to G ∗ .

2. Characters and duality 215Proof. In the first place Z is a finite group. For, if a∈G, ae, there isa characterχ such thatχ(a)1.If G is of order n, thenχ(a n )=(χ(a)) n = 1 so thatχ(a) in an elementof Z of finite order. Since Z is cyclic it follows that Z is finite. □From this it follows that G ∗ is finite.Since (χ(a)) n = 1 for everyχand every a, there is no loss in generalityif we assume that Z is a cyclic group of order n.In order to prove the theorem let us first assume that G is a cyclicgroup of order n. Let a be a generator of G and b a generator of thecyclic group Z of order n. Define the characterχ 1 of G by 245χ 1 (a)=b.Since a generates G any character is determined uniquely by its effecton a.χ 1 is an element of order n in G ∗ . Letχbe any character of G.Letχ(a)=b µfor some integerµ. Consider the character ˜χ=χ·χ −µ1 .˜χ(a)=(χ 1 (a)) −µ χ(a)=b −µ b µ = 1which shows that ˜χ=χ o is the principal character. Hence G ∗ is a cyclicgroup of order n and the mappinga→χ 1establishes an isomorphism of G on G ∗ .Let now G be finite non-cyclic abelian of order n. G is then a directproduct of cyclic groups G 1 ,..., G l of ordersλ 1 ,...,λ l respectively. Leta i be a generator of G i so that a 1 ,...,a l is a base of G. Sinceλ 1 ,...,λ ldivide n, we define l charactersχ 1 ,χ 2 ,...,χ l of G byχ i (a j )=1χ i (a i )=b ijii=1,...,l,

216 8. Appendix246247where b i is an element in Z of orderλ i . These characters are then independentelements of the abelian group G ∗ . For, ifχ t 11···χ t ll=χ 0 , then,for any a∈G,χ t 11(a)...χ t ll (a)=1.Taking for a successively a 1 ,...,a l we see thatλ i |t i and soχ 1 ,...,χ lare independent.Letχbe any character of G. Thenχis determined by its effect ona 1 ,...,a l . Letχ(a i )= s i . Since a λ ii= e, (χ(a i )) λ i= 1. But (χ(a i )) λ i=s λ ii. Thus s λ ii= 1. Z being cyclic, there exists only one subgroup oforderλ i . Thusχ(a i )= s i = b µ ii,for some integerµ i (modλ i ). Consider the character ˜χ=χχ −µ 11···χ. Itus clear that ˜χ(a i )=1 for all i so that ˜χ=χ o orµ µ1χ=χ 1···χ lThus G ∗ is the product of cyclic group generated byχ 1 ,...,χ 1 . ByCorollary to theorem 2, it follows that G and G ∗ are isomorphic.Corollary. If G is a finite group and G ∗ its character group, then whatevermay be Z,OrderG ∗ ≤ OrderG.Proof. For, if H is the subgroup of G defined earlier, then G/H is abelianand finite, since G is finite. Also Z is admissible for G/H. Furthermoreevery character of G can be considered as a character of G/H, by definitionof H. Hence by theorem 3 □Order G ∗ ≤ Order G/H≤ Order G.Let us now go back to the situation where G is finite abelian and Zia admissible for G. Then G≃ G ∗ . Let us define on G× G ∗ the functionl .(a,χ)=χ(a).For a fixedχ, the mapping a→(a,χ) is a character of G and so anelement of G ∗ . By definition of product of character, it follows that, forfixed a, the mappingā :χ→(a,χ)

3. Pairing of two groups 217is a homomorphism of G ∗ into Z and hence a character of G ∗ , Let G ∗∗denote the character group of G ∗ . By Corollary above,Consider now the mappingOrderG ∗∗ ≤ OrderG ∗ = OrderG.σ : a→āof G into G ∗∗ . This is clearly a homomorphism. If ā is identity, then(a,χ)=1 for allχ. But since Z is admissible for G, it follows that a=e.Henceσis an isomorphism of G into G ∗∗ . Therefore we haveTheorem 4. The mapping a→ā is a natural isomorphism of G ∗ on G ∗∗Note that the isomorphism of G on G ∗ is not natural.Under the conditions of theorem 3, we call G ∗ the dual of G. ThenG ∗∗ is the dual of G ∗ and theorem 4 shows that the dual of G ∗ is naturallyisomorphic to G. Theorem 4 is called the duality theorem for finiteabelian groups.3 Pairing of two groupsLet G and G ′ be two groups,σ,σ ′ ,..., elements of G ′ andτ,τ ′ ,...,elements of G ′ . Let Z be a cyclic group. Suppose there is a function 248(σ,τ) on G× G ′ into Z such that for everyσ, the mappingλ σ :τ→(σ,τ)is a homomorphism of G ′ into Z and for everyτ, the mappingµ τ :σ→(σ,τ)is a homomorphism of G into Z, Thusλ σ andµ τ are characters of G ′and G respectively. We then say that G and G ′ are paired to Z and that(σ,τ) is a pairing.For everyσ, let G ′ σ denote the kernel in G′ of the homomorphismλ σ . Put ⋂H ′ =G ′ σσ

218 8. Appendixfor allσ∈G. Then G ′ /H ′ is abelian. Also, H ′ is the set ofτin G ′ suchthat(σ,τ)=1for allσ∈G. Define in a similar way the subgroup H of G.We are going to proveTheorem 5. If G/H is finite, then so is G ′ /H ′ and both are then isomorphicto each other.Proof. From fixedσ, consider the functionχ σ (τ)=(σ,τ).This is clearly a character of G ′ . By definition of H ′ , it follows thatfor eachσ,χ σ is a character of G ′ /H ′ .Consider now the mappingσ→χ σof G into (G ′ /H ′ ) ∗ . This is again a homomorphism of G into (G ′ /H ′ ) ∗ .The kernel of the homomorphism is the set ofσfor whichχ σ is theprincipal character of G ′ /H ′ . By definition of H, it follows that H is thekernel. HenceG/H≃ a subgroup of (G ′ /H ′ ) ∗ . (1)□249In a similar way(G ′ /H ′ )≃ a subgroup of (G/H) ∗ . (2)Let now G/H be finite. Then by Corollary to theorem 3,By (2), this means thatord(G/H) ∗ ≤ ordG/H.order(G ′ /H ′ )≤ order G/H.Reversing the roles of G and G ′ we see that G/H and G ′ /H ′ havethe same order. (1) and (2) together with theorem 3 prove the theorem.If, in particular, we take G ∗ for G ′ , then H ′ = (χ o ) and so we have

3. Pairing of two groups 219Corollary. If G is any group, G ∗ its character group and if G/H is finitethenG/H≃ G ∗ .

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