INTRODUCTION TO ALGEBRAIC GEOMETRY Note del corso di ...

Introduction to algebraic geometry 7Let K[x 1 , . . ., x n ] be the polynomial ring in n variables over the field K. IfP(a 1 , . . ., a n ) ∈ A n , and F = F(x 1 , . . ., x n ) ∈ K[x 1 , . . ., x n ], we can consider thevalue of F at P, i.e. F(P) = F(a 1 , . . ., a n ) ∈ K. We say that P is a zero of F ifF(P) = 0.For example the points P 1 (1, 0), P 2 (−1, 0), P 3 (0, 1) are zeroes of F = x 2 + 1x 2 − 1 over any field. If G = 2 x2 + 1 x2 + 1 then G has no zeroes in 2 A2 , but it doesRhave zeroes in A 2 C.2.1. Definition. A subset X of A n K is an affine algebraic set if X is the set ofcommon zeroes of a family of polynomials of K[x 1 , . . ., x n ].This means that there exists a subset S ⊂ K[x 1 , . . ., x n ] such thatX = {P ∈ A n | F(P) = 0 ∀ F ∈ S}.In this case X is called the zeroes set of S and is denoted V (S) (or in some booksZ(S), e.g. this is the notation of Hartshorne’s book). In particular, if S = {F },then V (S) will be simply denoted by V (F).2.2. Examples and remarks.1. S = K[x 1 , . . ., x n ]: then V (S) = ∅, because S contains non–zero constants.2. S = {0}: then V (S) = A n .3. S = {xy − 1} : then V (xy − 1) is the hyperbola.4. If S ⊂ T, then V (S) ⊃ V (T).Let S ⊂ K[x 1 , . . ., x n ] be a set of polynomials, let α := 〈S〉 be the ideal generatedby S. Recall that α = {finite sums of products of the form HF where F ∈ S, H ∈K[x 1 , . . ., x n ]}.2.3. Proposition. V (S) = V (α).Proof. If P ∈ V (α), then F(P) = 0 for any F ∈ α; in particular for any F ∈ Sbecause S ⊂ α.Conversely, if P ∈ V (S), let G = ∑ i H iF i be a polynomial of α (F i ∈ S ∀ i).Then G(P) = ( ∑ H i F i )(P) = ∑ H i (P)F i (P) = 0.□The above Proposition is important in view of the following:Hilbert’ Basis Theorem. If R is a Noetherian ring, then the polynomial ringR[x] is Noetherian.Proof. See for instance [4].□2.4. Corollary. Any affine algebraic set X ⊂ A n is the zeroes set of a finitenumber of polynomials, i.e. there exist F 1 , . . ., F r ∈ K[x 1 , . . ., x n ] such that X =V (F 1 , . . ., F r ).□

8 MezzettiNote that V (F 1 , . . ., F r ) = V (F 1 )∩. . .∩V (F r ), so every algebraic set is a finiteintersection of algebraic sets of the form V (F), i.e. zeroes of a unique polynomialF. If F = 0, then V (0) = A n ; if F = c ∈ K \ {0}, then V (c) = ∅; if deg F > 0,then V (F) is called a hypersurface.2.5. Proposition. The affine algebraic sets of A n satisfy the axioms of theclosed sets of a topology, called the Zariski topology.Proof. It is enough to check that finite unions and arbitrary intersections of algebraicsets are again algebraic sets.Let V (α), V (β) be two algebraic sets, with α, β ideals of K[x 1 , . . ., x n ]. ThenV (α) ∪ V (β) = V (α ∩ β) = V (αβ), where αβ is the product ideal, defined by:αβ = { ∑ a i b i | a i ∈ α, b i ∈ β}.finIn fact: αβ ⊂ α ∩ β so V (α ∩ β) ⊂ V (αβ), and both α ∩ β ⊂ α and α ∩ β ⊂ βso V (α) ∪ V (β) ⊂ V (α ∩ β). Assume now that P ∈ V (αβ) and P /∈ V (α): hence∃F ∈ α such that F(P) ≠ 0; on the other hand, if G ∈ β then FG ∈ αβ so(FG)(P) = 0 = F(P)G(P), which implies G(P) = 0.Let V (α i ), i ∈ I, be a family of algebraic sets, α i ⊂ K[x 1 , . . ., x n ]. Then∩ i∈I V (α i ) = V ( ∑ i∈I α i), where ∑ i∈I α i is the sum ideal of α ′ s. In fact α i i∑⊂i∈I α i ∀i, hence V ( ∑ i α i) ⊂ V (α i ) ∀i and V ( ∑ i α i) ⊂ ∩ i V (α i ). Conversely, ifP ∈ V (α i ) ∀i, and F ∈ ∑ i α i, then F = ∑ F i i; therefore F(P) = ∑ F i (P) = 0.□2.6. Examples.1. The Zariski topology of the affine line A 1 .Let us recall that the polynomial ring K[x] in one variable is a PID (principalideal domain), so every ideal I ⊂ K[x] is of the form I = 〈F 〉. Hence every closedsubset of A 1 is of the form X = V (F), the set of zeroes of a unique polynomialF(x). If F = 0, then V (F) = A 1 , if F = c ∈ K ∗ , then V (F) = ∅, if deg F = d > 0,then F can be decomposed in linear factors in polynomial ring over the algebraicclosure of K; it follows that V (F) has at most d points.We conclude that the closed sets in the Zariski topology of A 1 are: A 1 , ∅ andthe finite sets.2. If K = R or C, then the Zariski topology and the usual topology on A ncan be compared. Every closed set in the Zariski topology is closed also in theusual topology: this is better seen working with open sets. Let X = V (F 1 , . . ., F r )be a closed set in the Zariski topology, and U := A n \ X; if P ∈ U, then ∃ F i suchthat F i (P) ≠ 0, so there exists an open neighbourhood of P in the usual topologyin which F i does not vanish.Conversely, there exist closed sets in the usual topology which are not Zariskiclosed, for example the balls. The first case, of an interval in the real affine line,follows from part 1.

Introduction to algebraic geometry 9We want to define now the projective algebraic sets in P n . Let K[x 0 , x 1 , . . ., x n ]be the polynomial ring in n + 1 variables. Fix a polynomial G(x 0 , x 1 , . . ., x n ) ∈K[x 0 , x 1 , . . ., x n ] and a point P[a 0 , a 1 , . . ., a n ] ∈ P n : then, in general,so the value of G at P is not defined.G(a 0 , . . ., a n ) ≠ G(λa 0 , . . ., λa n ),2.7. Example. Let G = x 1 +x 0 x 1 +x 2 2, P[0, 1, 2] = [0, 2, 4] ∈ P 2 R. So G(0, 1, 2) =1 + 4 ≠ G(0, 2, 4) = 2 + 16. But if Q = [1, 0, 0] = [λ, 0, 0], then G(1, 0, 0) =G(λ, 0, 0) = 0 for all λ.2.8. Definition. Let G ∈ K[x 0 , x 1 , . . ., x n ]: G is homogeneous of degree d if Gis a linear combination of monomials of degree d.2.9. Lemma. If G is homogeneous of degree d, G ∈ K[x 0 , x 1 , . . ., x n ], and t is anew variable, then G(tx 0 , . . ., tx n ) = t d G(x 0 , . . ., x n ).Proof. It is enough to prove the equality for monomials, i.e. forG = ax i 00 xi 11 . . .xi nn with i 0 + i 1 + . . . + i n = d :G(tx 0 , . . ., tx n ) = a(tx 0 ) i 0(tx 1 ) i 1. . .(tx n ) i n= at i 0+i 1 +...+i nx i 00x i 11. . .x i nn== t d G(x 0 , . . ., x n ).□2.10. Definition. Let G be a homogeneous polynomial of K[x 0 , x 1 , . . ., x n ]. Apoint P[a 0 , . . ., a n ] ∈ P n is a zero of G if G(a 0 , . . ., a n ) = 0. In this case we writeG(P) = 0.Note that by Lemma 2.9 if G(a 0 , . . ., a n ) = 0, thenfor every choice of λ ∈ K ∗ .G(λa 0 , . . ., λa n ) = λ deg G G(a 0 , . . ., a n ) = 02.11. Definition. A subset Z of P n is a projective algebraic set if Z is the setof common zeroes of a set of homogeneous polynomials of K[x 0 , x 1 , . . ., x n ].If T is such a subset of K[x 0 , x 1 , . . ., x n ], then the corresponding algebraic setwill be denoted by V P (T).Let α = 〈T 〉 be the ideal generated by the (homogeneous) polynomials of T.If F ∈ α, then F = ∑ i H iF i , F i ∈ T: if P ∈ V P (T), and P[a 0 , . . ., a n ], thenF(a 0 , . . ., a n ) = ∑ H i (a 0 , . . ., a n )F i (a 0 , . . ., a n ) = 0, for any choice of coordinatesof P, regardless if F is homogeneous or not. We say that P is a projective zero ofF.

10 MezzettiIf F is a polynomial, then F can be written in a unique way as a sum ofhomogeneous polynomials, called the homogeneous components of F: F = F 0 +F 1 + . . . + F d . More in general, we give the following:2.12. Definition. Let A be a ring. A is called a graded ring over Z if there existsa family of additive subgroups {A i } i∈Z such that A = ⊕ i∈Z A i and A i A j ⊂ A i+jfor all pair of indices.The elements of A i are called homogeneous of degree i and A i is the homogeneouscomponent of degree i. The standard example of graded ring is the polynomialring with coefficients in a ring R. In this case the homogeneous componentsof negative degrees are all zero.2.13 Proposition - Definition. Let I ⊂ A be an ideal of a graded ring. I iscalled homogeneous if the following equivalent conditions are fulfilled:(i) I is generated by homogeneous elements;(ii) I = ⊕ k∈Z (I ∩A k ), i.e. if F = Σ k∈Z F k ∈ I, then all homogeneous componentsF k of F belong to I.Proof of the equivalence.“(ii)⇒(i)”: given a system of generators of I, write each of them as sum of itshomogeneous components: F i = Σ k∈Z F ik . Then a set of homogeneous generatorsof I is formed by all the elements F ik .“(i)⇒(ii)”: let I be generated by a family of homogeneous elements {G α },with deg G α = d α . If F ∈ I, then F is a combination of the elements G α withsuitable coefficients H α ; write each H α as sum of its homogeneous components:H α = ΣH αk . Note that the product H αk G α is homogeneous of degree k + d α . Bythe unicity of the expression of F as sum of homogeneous elements, it follows thatall of them are combinations of the generators {G α } and therefore they belong toI. □Let I ⊂ K[x 0 , x 1 , . . ., x n ] be a homogeneous ideal. Note that, by the noetherianity,I admits a finite set of homogeneous generators.Let P[a 0 , . . ., a n ] ∈ P n . If F ∈ I, F = F 0 + . . . + F d , then F 0 ∈ I, . . ., F d ∈ I.We say that P is a zero of I if P is a projective zero of any polynomial of I or,equivalently, of any homogeneous polynomial of I. This also means that P is azero of any homogeneous polynomial of a set generating I. The set of zeroes of Iwill be denoted V P (I): all projective algebraic subsets of P n are of this form.As in the affine case, the projective algebraic subsets of P n satisfy the axiomsof the closed sets of a topology called the Zariski topology of P n (see also Exercise3).Note that also all subsets of A n and P n have a structure of topological space,with the induced topology, which is still called the Zariski topology.Exercises to §2.

Introduction to algebraic geometry 111. Let F ∈ K[x 1 , . . ., x n ] be a non–constant polynomial. The set A n \ V (F)will be denoted A n F . Prove that {An F |F ∈ K[x 1, . . ., x n ] \ K} is a topology basisfor the Zariski topology.2. Let B ⊂ R n be a ball. Prove that B is not Zariski closed.3*. Let I, J be homogeneous ideals of K[x 0 , x 1 , . . ., x n ]. Prove that I + J, IJand I ∩ J are homogeneous ideals.4*. Prove that the map φ : A 1 → A 3 defined by t → (t, t 2 , t 3 ) is a homeomorphismbetween A 1 and its image, for the Zariski topology.5. Let X ⊂ A 2 Rbe the graph of the map R → R such that x → sin x. Is Xclosed in the Zariski topology? (hint: intersect X with a line....)3. Examples of algebraic sets.a) In the Zariski topology both of A n and of P n all points are closed.If P(a 1 , . . ., a n ) ∈ A n : P = V (x 1 − a 1 , . . ., x n − a n ). If P[a 0 , . . ., a n ] ∈ P n :P = V P (〈a i x j − a j x i 〉 i,j=0,...,n ).Note that in the projective case the polynomials defining P as closed set arehomogeneous. They can be seen as minors of order 2 of the matrixwith entries in K[x 0 , x 1 , . . ., x n ].( )a0 a 1 . . . a nx 0 x 1 . . . x nb) Hypersurfaces.Let us recall that the polynomial ring K[x 1 , . . ., x n ] is a UFD (unique factorizationdomain), i. e. every non-constant polynomial F can be expressed ina unique way (up to the order and up to units) as F = F r 11F r 22. . .F r ss, whereF 1 , . . ., F s are irreducible polynomials, two by two distinct, and r i ≥ 1∀ i =1, . . ., s. Hence the hypersurface of A n defined by F isX := V (F) = V (F r 11F r 22. . .F r ss) = V (F 1 F 2 . . .F s ) = V (F 1 ) ∪ V (F 2 ) ∪ . . . ∪ V (F s ).The equation F 1 F 2 . . .F s = 0 is called the reduced equation of X. Notethat F 1 F 2 . . .F s generates the radical √ F. If s = 1, X is called an irreduciblehypersurface; by definition its degree is the degree of its reduced equation. Anyhypersurface is a finite union of irreducible hypersurfaces.In a similar way one defines hypersurfaces of P n , i. e. projective algebraicsets of the form Z = V P (G), with G ∈ K[x 0 , x 1 , . . ., x n ], G homogeneous. Sincethe irreducible factors of G are homogeneous (see Exercise 3.6), any projectivehypersurface Z has a reduced equation (whose degree is, by definition, the degreeof Z) and Z is a finite union of irreducible hypersurfaces. The degree of a projectivehypersurface has the following important geometrical meaning.

12 Mezzetti3.1. Proposition. Let K be an algebraically closed field. Let Z ⊂ P n be aprojective hypersurface of degree d. Then a line of P n , not contained in Z, meetsZ at exactly d points, counting multiplicities.Proof. Let G be the reduced equation of Z and L ⊂ P n be any line.We fix two points on L: A = [a 0 , . . ., a n ], B = [b 0 , . . ., b n ]. So L admitsparametric equations of the form⎧x 0 = λa 0 + µb 0⎪⎨x 1 = λa 1 + µb 1⎪⎩ . . .x n = λa n + µb nThe points of Z ∩L are obtained from the homogeneous pairs [λ, µ] which aresolutions of the equation G(λa 0 + µb 0 , . . ., λa n + µb n ) = 0. If L ⊂ Z, then thisequation is identical. Otherwise, G(λa 0 + µb 0 , . . ., λa n + µb n ) is a non-zero homogeneouspolynomial of degree d in two variables. Being K algebraically closed, itcan be factorized in linear factors:G(λa 0 + µb 0 , . . ., λa n + µb n ) = (µ 1 λ − λ 1 µ) d 1(µ 2 λ − λ 2 µ) d 2. . .(µ r λ − λ r µ) d rwith d 1 + d 2 + . . . + d r = d. Every factor corresponds to a point in Z ∩ L, to becounted with the same multiplicity as the factor.□If K is not algebraically closed, considering the algebraic closure of K andusing Proposition 3.1, we get that d is un upper bound on the number of pointsof Z ∩ L.c) Affine and projective subspaces.The subspaces introduced in §1, both in the affine and in the projective case,are examples of algebraic sets.d) Product of affine spaces.Let A n , A m be two affine spaces over the field K. The cartesian productA n ×A m is the set of pairs (P, Q), P ∈ A n , Q ∈ A m : it is in natural bijection withA n+m via the mapφ : A n × A m −→ A n+msuch that φ((a 1 , . . ., a n ), (b 1 , . . ., b m )) = (a 1 , . . ., a n , b 1 , . . ., b m ).From now on we will always identify A n × A m with A n+m . We get twotopologies on A n × A m : the Zariski topology and the product topology.3.1. Proposition. The Zariski topology is strictly finer than the product topology.Proof. If X = V (α) ⊂ A n , α ⊂ K[x 1 , . . ., x n ] and Y = V (β) ⊂ A m , β ⊂K[y 1 , . . ., y m ], then X × Y ⊂ A n × A m is Zariski closed, precisely X × Y =

Introduction to algebraic geometry 13V (α ∪ β) where the union is made in the polynomial ring in n + m variablesK[x 1 , . . ., x n , y 1 , . . ., y m ]. Hence, if U = A n \ X, V = A m \ Y are open subsets ofA n and A m in the Zariski topology, then U ×V = A n ×A m \((A n ×Y )∪(X ×A m ))is open in A n × A m in the Zariski topology.Conversely, we prove that A 1 × A 1 = A 2 contains some subsets which areZariski open but are not open in the product topology. The proper open subsetsin the product topology are of the form A 1 × A 1 \ { finite unions of “vertical” and“ horizontal” lines}.– Fig.1 –Let X = A 2 \V (x −y): it is Zariski open but does not contain any non-emptysubset of the above form, so it is not open in the product topology. There aresimilar examples in A n × A m for any n, m.□Note that there is no similar construction for P n × P m .e) Embedding of A n in P n .Let H i be the hyperplane of P n of equation x i = 0, i = 0, . . ., n; it is closedin the Zariski topology, and the complementar set U i is open. So we have an opencovering of P n : P n = U 0 ∪ U 1 ∪ . . . ∪ U n . Let us recall that for all i there is abijection φ i : U i → A n such that φ i ([x 0 , . . ., x i , . . ., x n ]) = ( x 0x i, . . ., ˆ1, . . ., x nxi). Theinverse map is j i : A n → U i such that j i (y 1 , . . ., y n ) = [y 1 , . . ., 1, . . ., y n ].3.2. Proposition. The map φ i is a homeomorphism, for i = 0, . . ., n.Proof. Assume i = 0 (the other cases are similar).We introduce two maps:(i) dehomogeneization of polynomials with respect to x 0 .

14 MezzettiIt is a map a : K[x 0 , x 1 , . . ., x n ] → K[y 1 , . . ., y n ] such thata (F(x 0 , . . ., x n )) = a F(y 1 , . . ., y n ) := F(1, y 1 , . . ., y n ).Note that a is a ring homomorphism.(ii) homogeneization of polynomials with respect to x 0 .It is a map h : K[y 1 , . . ., y n ] → K[x 0 , x 1 , . . ., x n ] defined by.h (G(y 1 , . . ., y n )) = h G(x 0 , . . ., x n ) := x deg G0G( x 1x 0, . . ., x nx 0)h G is always a homogeneous polynomial of the same degree as G. The map his clearly not a ring homomorphism. Note that always a ( h G) = G but in generalh ( a F) ≠ F; what we can say is that, if F(x 0 , . . ., x n ) is homogeneous, then ∃r ≥ 0such that F = x r 0( h ( a F)).Let X ⊂ U 0 be closed in the topology induced by the Zariski topology ofthe projective space, i.e. X = U 0 ∩ V P (I) where I is a homogeneous ideal ofK[x 0 , x 1 , . . ., x n ]. Define a I = { a F | F ∈ I}: it is an ideal of K[y 1 , . . ., y n ](because a is a ring homomorphism). We prove that φ 0 (X) = V ( a I). For:let P[x 0 , . . ., x n ] be a point of U 0 ; then φ 0 (P) = ( x 1x 0, . . ., x nx0) ∈ φ 0 (X) ⇐⇒P[x 0 , . . ., x n ] = [1, x 1x 0, . . ., x nx0] ∈ X = V P (I) ⇐⇒ F(1, x 1x 0, . . ., x nx0) = 0 ∀ a F ∈a I ⇐⇒ φ 0 (P) ∈ V ( a I).Conversely: let Y = V (α), α ideal of K[y 1 , . . ., y n ], be a Zariski closed setof A n . Let h α be the homogeneous ideal of K[x 0 , x 1 , . . ., x n ] generated by the set{ h G | G ∈ α}. We prove that φ −10(Y ) = V P ( h α) ∩ U 0 . In fact: [1, x 0 , . . ., x n ] ∈φ −1 (Y ) ⇐⇒ (x0 1 , . . ., x n ) ∈ Y ⇐⇒ G(x 1 , . . ., x n ) = h G(1, x 1 , . . ., x n ) = 0 ∀ G ∈α ⇐⇒ [1, x 1 , . . ., x n ] ∈ V P ( h α).□From now on we will often identify A n with U 0 via φ 0 (and similarly with U ivia φ i ). So if P[x 0 , . . ., x n ] ∈ U 0 , we will refer to x 0 , . . ., x n as the homogeneouscoordinates of P and to x 1x 0, . . ., x nx0as the non–homogeneous or affine coordinatesof P.Exercises to §3.1*. Let n ≥ 2. Prove that, if K is an algebraically closed field, then in A n Kboth any hypersurface and any complementar set of a hypersurface have infinitelymany points.2. Prove that the Zariski topology on A n is T 1 .3*. Let F ∈ K[x 0 , x 1 , . . ., x n ] be a homogeneous polynomial. Check that itsirreducible factors are homogeneous. (hint: consider a product of two polynomialsnot both homogeneous...)4. The ideal of an algebraic set and the Hilbert Nullstellensatz.

Introduction to algebraic geometry 15Let X ⊂ A n be an algebraic set, X = V (α), α ⊂ K[x 1 , . . ., x n ]. The ideal αdefining X is not unique: for example, let X = {0} ⊂ A 2 ; then 0 = V (x 1 , x 2 ) =V (x 2 1, x 2 ) = V (x 2 1, x 3 2) = V (x 2 1, x 1 , x 2 , x 2 2) = . . . Nevertheless, there is an ideal wecan canonically associate to X, i.e. the biggest one. Precisely:4.1. Definition. Let Y ⊂ A n be any set.The ideal of Y is I(Y ) = {F ∈ K[x 1 , . . ., x n ] | F(P) = 0 for any P ∈ Y } ={F ∈ K[x 1 , . . ., x n ] | Y ⊂ V (F)}: it is formed by all polynomials vanishing on Y .Note that I(Y ) is in fact an ideal.For instance, if P(a 1 , . . ., a n ) is a point, then I(P) = 〈x 1 − a 1 , . . ., x n − a n 〉.Indeed all its polynomials vanish on P, and, on the other side, it is maximal.The following relations follow immediately by the definition:(i) if Y ⊂ Y ′ , then I(Y ) ⊃ I(Y ′ );(ii) I(Y ∪ Y ′ ) = I(Y ) ∩ I(Y ′ );(iii) I(Y ∩ Y ′ ) ⊃ I(Y ) + I(Y ′ ).Similarly, if Z ⊂ P n is any set, the homogeneous ideal of Z is, by definition,the homogeneous ideal of K[x 0 , x 1 , . . ., x n ] generated by the set {G ∈K[x 0 , x 1 , . . ., x n ] | G is homogeneous and V P (G) ⊃ Z}. It is denoted I h (Z).Relations similar to (i),(ii),(iii) are satisfied. I h (Z) is also the set of polynomialsF(x 0 , . . ., x n ) such that every point of Z is a projective zero of F.Let α ⊂ K[x 1 , . . ., x n ] be an ideal. Let √ α denote the radical of α, i.e. theideal {F ∈ K[x 1 , . . ., x n ] | ∃r ≥ 1 s.t. F r ∈ α}. Note that always α ⊂ √ α; ifequality holds, then α is called a radical ideal.4.2. Proposition.1) For any X ⊂ A n , I(X) is a radical ideal.2) For any Z ⊂ P n , I h (Z) is a homogeneous radical ideal.Proof. 1) If F ∈ √ I(X), let r ≥ 1 such that F r ∈ I(X): hence if P ∈ X, then(F r )(P) = 0 = (F(P)) r in the base field K. Therefore F(P) = 0.2) is similar, taking into account that I h (Z) is a homogeneous ideal (seeExercise 4.7.).□We can interpret I as a map from P(A n ), the set of subsets of the affine space,to P(K[x 1 , . . ., x n ]). On the other hand, V can be seen as a map in the oppositesense. We have:4.3. Proposition. Let α ⊂ K[x 1 , . . ., x n ] be an ideal, Y ⊂ A n be any subset.Then:(i) α ⊂ I(V (α));(ii) Y ⊂ V (I(Y ));(iii) V (I(Y )) = Y : the closure of Y in the Zariski topology of A n .

16 MezzettiProof. (i) If F ∈ α and P ∈ V (α), then F(P) = 0, so F ∈ I(V (α)).(ii) If P ∈ Y and F ∈ I(Y ), then F(P) = 0, so P ∈ V (I(Y )).(iii) Taking closures in (ii), we get: Y ⊂ V (I(Y )) = V (I(Y )). Conversely,let X = V (β) be any closed set containing Y : X = V (β) ⊃ Y . Then I(Y ) ⊃I(V (β)) ⊃ β by (i); we apply V again: V (β) = X ⊃ V (I(Y )) so any closed setcontaining Y contains V (I(Y )) so Y ⊃ V (I(Y )).□Similar properties relate homogeneous ideals of K[x 0 , x 1 , . . ., x n ] and subsetsof P n ; in particular, if Z ⊂ P n , then V P (I h (Z)) = Z, the closure of Z in the Zariskitopology of P n .There does not exist any characterization of I(V (α)) in general. We can onlysay that it is a radical ideal containing α, so it contains also √ α. To characterizeI(V (α)) we need some extra assumption on the base field.4.4. Hilbert Nullstellensatz (Theorem of zeroes). Let K be an algebraicallyclosed field. Let α ⊂ K[x 1 , . . ., x n ] be an ideal. Then I(V (α)) = √ α.The assumption on K is necessary. Let me recall that K is algebraicallyclosed if any non–constant polynomial of K[x] has at least one root in K, or,equivalently, if any irreducible polynomial of K[x] has degree 1. So if K is notalgebraically closed, there exists F ∈ K[x], irreducible of degree d > 1. ThereforeF has no zero in K, hence V (F) ⊂ A 1 K is empty. So I(V (F)) = I(∅) = {G ∈K[x] | ∅ ⊂ V (G)} = K[x]. But 〈F 〉 is a maximal ideal of K[x], and 〈F 〉 ⊂ √ 〈F 〉.If 〈F 〉 ≠ √ 〈F 〉, by the maximality √ 〈F 〉 = 〈1〉, so ∃r ≥ 1 such that 1 r = 1 ∈ 〈F 〉,which is false. Hence √ 〈F 〉 = 〈F 〉 ≠ K[x] = I(V (F)).Proof. of the Nullstellensatz. It will follow from two facts:1 st fact. Let K be an algebraically closed field, let M ⊂ K[x 1 , . . ., x n ] be amaximal ideal. Then, there exist a 1 , . . ., a n ∈ K such that M = 〈x 1 −a 1 , . . ., x n −a n 〉.2 nd fact (Weak Nullstellensatz).Let K be an algebraically closed field, let α ⊂ K[x 1 , . . ., x n ] be a proper ideal.Then V (α) ≠ ∅ i.e. the polynomials of α have at least one common zero in A n K .Indeed, since α is proper, there exists a maximal ideal M containing α. ThenV (α) ⊃ V (M). By 1 st Step, M = 〈x 1 − a 1 , . . ., x n − a n 〉, so V (M) = {P } withP(a 1 , . . ., a n ), hence P ∈ V (α).3 rd Step (Rabinowitch method).Let K be an algebraically closed field: we will prove that I(V (α)) ⊂ √ α.Since the reverse inclusion always holds, this will conclude the proof.Let F ∈ I(V (α)), F ≠ 0 and let α = 〈G 1 , . . ., G r 〉. The assumption onF means: if G 1 (P) = . . . = G r (P) = 0, then F(P) = 0. Let us consider thepolynomial ring in n + 1 variables K[x 1 , . . ., x n+1 ] and let β be the ideal β =

Introduction to algebraic geometry 17〈G 1 , . . ., G r , x n+1 F − 1〉: β has no zeroes in A n+1 , hence, by Step 1, 1 ∈ β, i.e.there exist H 1 , . . ., H r+1 ∈ K[x 1 , . . ., x n+1 ] such that1 = H 1 G 1 + . . . + H r G r + H r+1 (x n+1 F − 1).We introduce the homomorphism ψ : K[x 1 , . . ., x n+1 ] → K(x 1 , . . ., x n ) definedby H(x 1 , . . ., x n+1 ) → H(x 1 , . . ., x n , 1 F ).The polynomials G 1 , . . ., G r do not contain x n+1 so ψ(G i ) = G i ∀ i = 1, . . ., r.Moreover ψ(x n+1 F − 1) = 0, ψ(1) = 1. Therefore1 = ψ(H 1 G 1 + . . . + H r G r + H r+1 (x n+1 F − 1)) = ψ(H 1 )G 1 + . . . + ψ(H r )G rwhere ψ(H i ) is a rational function with denominator a power of F. By multiplyingthis relation by a common denominator, we get an expression of the form:F m = H ′ 1G 1 + . . . + H ′ rG r ,so F ∈ √ α.□4.10. Corollaries. Let K be an algebraically closed field.1. There is a bijection between algebraic subsets of A n and radical ideals ofK[x 1 , . . ., x n ]. The bijection is given by α → V (α) and X → I(X). In fact, if Xis closed in the Zariski topology, then V (I(X)) = X; if α is a radical ideal, thenI(V (α)) = α.2. Let X, Y ⊂ A n be closed sets. Then(i) I(X ∩ Y ) = √ I(X) + I(Y );(ii) I(X ∪ Y ) = I(X) ∩ I(Y ) = √ I(X)I(Y ).Proof. 2. follows from next lemma, using the Nullstellensatz.4.11. Lemma. Let α, β be ideals of K[x 1 , . . ., x n ]. Thena) √ √ α =√ α;b) √ α + β = √ √ α +√ β;c) √ α ∩ β = √ αβ = √ α ∩ √ β.Proof.a) if F ∈ √ √ α, there exists r ≥ 1 such that F r ∈ √ α, hence there exists s ≥ 1such that F rs ∈ α.b) α ⊂ √ α, β ⊂ √ β imply α + β ⊂ √ α + √ β hence √ α + β ⊂ √ √ √ α + β.Conversely, α ⊂ α + β, β ⊂ α + β imply √ α ⊂ √ α + β, √ β ⊂ √ α + β, hence√ √ √ √√ √ √√ √ α + β ⊂ α + β so α + β ⊂ α + β = α + β.c) αβ ⊂ α ∩ β ⊂ α (resp. ⊂ β) therefore √ αβ ⊂ √ α ∩ β ⊂ √ α ∩ √ β. IfF ∈ √ α ∩ √ β, then F r ∈ α, F s ∈ β for suitable r, s ≥ 1, hence F r+s ∈ αβ, soF ∈ √ αβ.□

18 MezzettiPart 2.(i) of 4.10. implies that, if I(X ∩Y ) ≠ I(X)+I(Y ), then I(X)+I(Y )is not radical.We move now to projective space. There exist proper homogeneous ideals ofK[x 0 , x 1 , . . ., x n ] without zeroes in P n , also assuming K algebraically closed: forexample the maximal ideal 〈x 0 , x 1 , . . ., x n 〉. The following characterization holds:4.12. Proposition. Let K be an algebraically closed field and let I be ahomogeneous ideal of K[x 0 , x 1 , . . ., x n ].The following are equivalent:(i) V P (I) = ∅;(ii) either I = K[x 0 , x 1 , . . ., x n ] or √ I = 〈x 0 , x 1 , . . ., x n 〉;(iii) ∃d ≥ 1 such that I ⊃ K[x 0 , x 1 , . . ., x n ] d , the subgroup of K[x 0 , x 1 , . . ., x n ]formed by the homogeneous polynomials of degree d.Proof.(i)⇒(ii) Let p : A n+1 − {0} → P n be the canonical surjection. We have:V P (I) = p(V (I)−{0}), where V (I) ⊂ A n+1 . So if V P (I) = ∅, then either V (I) = ∅or V (I) = {0}. If V (I) = ∅ then I(V (I)) = I(∅) = K[x 0 , x 1 , . . ., x n ]; if V (I) = {0},then I(V (I)) = 〈x 0 , x 1 , . . ., x n 〉 = √ I by the Nullstellensatz.(ii)⇒(iii) Let √ I = K[x 0 , x 1 , . . ., x n ], then 1 ∈ √ I so 1 r = 1 ∈ I(r ≥ 1). If√I = 〈x0 , x 1 , . . ., x n 〉, then for all variable x k there exists an index i k ≥ 1 suchthat x i kk∈ I. If d ≥ i 0 + i 1 + . . . + i n , then any monomial of degree d is in I, soK[x 0 , x 1 , . . ., x n ] d ⊂ I.(iii)⇒(i) because no point in P n has all coordinates equal to 0.□4.13. Theorem. Let K be an algebraically closed field and I be a homogeneousideal of K[x 0 , x 1 , . . ., x n ]. If F is a homogeneous non–constant polynomial suchthat V P (F) ⊃ V P (I) (i.e. F vanishes on V P (I), then F ∈ √ I.Proof. We have p(V (I) − {0}) = V P (I) ⊂ V P (F). Since F is non–constant, wehave also V (F) = p −1 (V P (F)) ∪ {0}, so V (F) ⊃ V (I); by the NullstellensatzI(V (I)) = √ I ⊃ I(V (F)) = √ (F) ∋ F.□4.14. Corollary (homogeneous Nullstellensatz). Let I be a homogeneousideal of K[x 0 , x 1 , . . ., x n ] such that V P (I) ≠ ∅, K algebraically closed. Then √ I =I h (V P (I)).□4.15. Definition. A homogeneous ideal of K[x 0 , x 1 , . . ., x n ] such that √ I =〈x 0 , x 1 , . . ., x n 〉 is called irrelevant.4.16. Corollary. Let K be an algebraically closed field. There is a bijectionbetween the set of projective algebraic subsets of P n and the set of radical

Introduction to algebraic geometry 19homogeneous non–irrelevant ideals of K[x 0 , x 1 , . . ., x n ].□Remark. Let X ⊂ P n be an algebraic set, X ≠ ∅. The affine cone ofX, denoted C(X), is the following subset of A n+1 : C(X) = p −1 (X) ∪ {0}. IfX = V P (F 1 , . . ., F r ), with F 1 , . . ., F r homogeneous, then C(X) = V (F 1 , . . ., F r ).By the Nullstellensatz, if K is algebraically closed, I(C(X)) = I h (X).Exercises to §4.1. Give a non-trivial example of an ideal α of K[x 1 , . . ., x n ] such that α ≠ √ α.2. Show that the following closed subsets of the affine plane Y = V (x 2 +y 2 −1)and Y ′ = V (y − 1) are such that equality does not hold in the following relation:I(Y ∩ Y ′ ) ⊃ I(Y ) + I(Y ′ ).3. Let α ⊂ K[x 1 , . . ., x n ] be an ideal. Prove that α = √ α if and only if thequotient ring K[x 1 , . . ., x n ]/α does not contain non–zero nilpotents.4*. Let I be a homogeneous ideal of K[x 1 , . . ., x n ] satisfying the followingcondition: if F is a homogeneous polynomial such that F r ∈ I for some positiveinteger r, then F ∈ I. Prove that I is a radical ideal.5. The projective closure of an affine algebraic set.Let X ⊂ A n be Zariski closed. Fix an index i ∈ {0, . . ., n} and embed A n into P nas the open subset U i . So X ⊂ A n φ i֒→ P n .5.1. Definition. The projective closure of X, X, is the closure of X in theZariski topology of P n .Since the map φ i is a homeomorphism (see Proposition 3.2.), we have: X ∩A n = X because X is closed in A n . The points of X ∩ H i , where H i = V P (x i ),are called the “points at infinity” of X in the fixed embedding.Note that, if K is an infinite field, then the projective closure of A n is P n :indeed, let F be a homogeneous polynomial vanishing along A n = U 0 . We canwrite F = F 0 x d 0 +F 1x0 d−1 +. . .+F d . By assumption, for every P(a 1 , . . ., a n ) ∈ A n ,P ∈ V P (F), i.e. F(1, a 1 , . . ., a n ) = 0 = a F(a 1 , . . ., a n ). So a F ∈ I(A n ). Weclaim that I(A n ) = (0): if n = 1, this follows from the principle of identity ofpolynomials, because K is infinite. If n ≥ 2, assume that F(a 1 , ..., a n ) = 0 forall (a 1 , ..., a n ) ∈ K n and consider F(a 1 , ..., a n−1 , x): either it has positive degreein x for some choice of (a 1 , ..., a n ), but then it has finitely many zeroes againstthe assumption; or it is always constant in x, so F belongs to K[x 1 , ..., x n−1 ] andwe can conclude by induction. So the claim is proved. We get therefore thatF 0 = F 1 = . . . = F d = 0 and F = 0.

20 Mezzetti5.2. Proposition. Let X ⊂ A n be an affine algebraic set, X be the projectiveclosure of X. ThenI h (X) = h I(X) := 〈 h F |F ∈ I(X)〉.Proof. Assume A n = U 0 ⊂ P n .Let F ∈ I h (X) be a homogeneous polynomial. If P(a 1 , . . ., a n ) ∈ X, then[1, a 1 , . . ., a n ] ∈ X, so F(1, a 1 , . . ., a n ) = 0 = a F(a 1 , . . ., a n ). Hence a F ∈ X.There exists k ≥ 0 such that F = (x k 0 )h ( a F) (see Proposition 3.2), so F ∈ h I(X).Hence I h (X) ⊂ h I(X).Conversely, if G ∈ I(X) and P(a 1 , . . ., a n ) ∈ X, then G(a 1 , . . ., a n ) = 0 =h G(1, a 1 , . . ., a n ), so h G ∈ I h (X) (here X is seen as a subset of P n ). So h I(X) ⊂I h (X). Since I h (X) = I h (X) (see Exercise 5.1), we have the claim.□In particular, if X is a hypersurface and I(X) = 〈F 〉, then I h (X) = 〈 h F 〉.Next example will show that, in general, it is not true that, if I(X) =〈F 1 , . . ., F r 〉, then h I(X) = 〈 h F 1 , . . ., h F r 〉. Only in the last twenty years, thanksto the development of symbolic algebra and in particular of the theory of Groebnerbases, the problem of characterizing the systems of generators of I(X), whosehomogeneization generates h I(X), has been solved.5.3. Example. The skew cubic.Let K be an algebraically closed field. The affine skew cubic is the followingclosed subset of A 3 : X = V (y − x 2 , z − x 3 ) (we use variables x, y, z). X is theimage of the map φ : A 1 → A 3 such that φ(t) = (t, t 2 , t 3 ). Note that φ : A 1 → Xis a homeomorphism (see Exercise 2.4). The ideal α = 〈y − x 2 , y − x 3 〉 defines Xand is prime: indeed the quotient ring K[x, y, z]/α is isomorphic to K[x], hencean integral domain. Therefore α is radical so α = I(X).Let X be the projective closure of X in P 3 . We are going to prove that Xis the image of the map ψ : P 1 → P 3 such that ψ([λ, µ]) = [λ 3 , λ 2 µ, λµ 2 , µ 3 ]. Weidentify A 1 with the open subset of P 1 defined by λ ≠ 0 i.e. U 0 , and A 3 withthe open subset of P 3 defined by x 0 ≠ 0 (U 0 too). Note that ψ| A1 = φ, becauseψ([1, t]) = [1, t, t 2 , t 3 ] = via the identification of A 3 with U 0 = (t, t 2 , t 3 ) = φ(t).Moreover ψ([0, 1]) = [0, 0, 0, 1]. So ψ(P 1 ) = X ∪ {[0, 0, 0, 1]}.If G is a homogeneous polynomial of K[x 0 , x 1 , . . ., x 3 ] such that X ⊂ V P (G),then G(1, t, t 2 , t 3 ) = 0 ∀t ∈ K, so G(λ 3 , λ 2 µ, λµ 2 , µ 3 ) = 0 ∀µ ∈ K, ∀λ ∈ K ∗ .Since K is infinite, then G(λ 3 , λ 2 µ, λµ 2 , µ 3 ) is the zero polynomial in λ and µ, soG(0, 0, 0, 1) = 0 and V P (G) ⊃ ψ(P 1 ), therefore X ⊃ ψ(P 1 ).Conversely, it is easy to prove that ψ(P 1 ) is Zariski closed, in fact that ψ(P 1 ) =V P (x 2 1− x 0 x 2 , x 1 x 2 − x 0 x 3 , x 2 2− x 1 x 3 ). So ψ(P 1 ) = X.

Introduction to algebraic geometry 21The three polynomials F 0 := x 1 x 3 − x 2 2, F 1 := x 1 x 2 − x 0 x 3 , F 2 := x 0 x 2 − x 2 1are the 2 × 2 minors of the matrix( )x0 xM =1 x 2x 1 x 2 x 3with entries in K[x 0 , x 1 , . . ., x 3 ]. Let F = y −x 2 , G = z −x 3 be the two generatorsof I(X); h F = x 0 x 2 −x 2 1, h G = x 2 0x 3 −x 3 1, hence V P ( h F, h G) = V P (x 0 x 2 −x 2 1, x 2 0x 3 −x 3 1 ) ≠ X, because V P( h F, h G) contains the whole line V P (x 0 , x 1 ).We shall prove now the non-trivial fact:5.4. Proposition. I h (X) = 〈F 0 , F 1 , F 2 〉.Proof. For all integer number d ≥ 0, let I h (X) d := I h (X) ∩ K[x 0 , x 1 , . . ., x 3 ] d : itis a K-vector space of dimension ≤ ( )d+33 . We define a K-linear map ρd havingI h (X) d as kernel:ρ d : K[x 0 , x 1 , . . ., x 3 ] d → K[λ, µ] 3dsuch that ρ d (F) = F(λ 3 , λ 2 µ, λ 2 µ 2 , µ 3 ). Since ρ d is clearly surjective, we compute( ) d + 3dim I h (X) d = − (3d + 1) = (d 3 + 6d 2 − 7d)/6.3For d ≥ 2, we define now a second K-linear mapφ d : K[x 0 , x 1 , . . ., x 3 ] d−2 ⊕ K[x 0 , x 1 , . . ., x 3 ] d−2 ⊕ K[x 0 , x 1 , . . ., x 3 ] d−2 → I h (X) dsuch that φ d (G 0 , G 1 , G 2 ) = G 0 F 0 + G 1 F 1 + G 2 F 2 . Our aim is to prove that φ d issurjective. The elements of its kernel are called the syzygies of degree d amongthe polynomials F 0 , F 1 , F 2 . Two obvious syzygies of degree 3 are constructed bydeveloping, according to the Laplace rule, the determinant of the matrix obtainedrepeating one of the rows of M, for example⎛⎝ x ⎞0 x 1 x 2x 0 x 1 x 2⎠x 1 x 2 x 3We put H 1 = (x 0 , x 1 , x 2 ) and H 2 = (x 1 , x 2 , x 3 ), they both belong to ker φ 3 . Notethat H 1 and H 2 give raise to syzygies of all degrees ≥ 3, in fact we can constructa third linear mapψ d : K[x 0 , x 1 , . . ., x 3 ] d−3 ⊕ K[x 0 , x 1 , . . ., x 3 ] d−3 → kerφ dputting ψ d (A, B) = H 1 A + H 2 B = (x 0 , x 1 , x 2 )A + (x 1 , x 2 , x 3 )B.

22 MezzettiClaim. ψ d is an isomorphism.Assuming the claim, we are able to compute dim ker φ d = 2 ( d3), therefore( ) d + 1dim Im φ d = 33( d− 23)which coincides with the dimension of I h (X) d previously computed. This provesthat φ d is surjective for all d and concludes the proof of the Proposition.Proof of the Claim. Let (G 0 , G 1 , G 2 ) belong to kerφ d . This means that thefollowing matrix N with entries in K[x 0 , x 1 , . . ., x 3 ] is degenerate:N :=⎛⎝ G ⎞0 G 1 G 2x 0 x 1 x 2⎠x 1 x 2 x 3Therefore, the rows of N are linearly dependent over the quotient field of thepolynomial ring K(x 0 , . . ., x 3 ). Since the last two rows are independent, thereexist reduced rational functions a 1a 0, b 1b 0∈ K(x 0 , x 1 , x 2 , x 3 ), such thatand similarlyG 0 = a 1a 0x 0 + b 1b 0x 1 = a 1b 0 x 0 + a 0 b 1 x 1a 0 b 0G 1 = a 1b 0 x 1 + a 0 b 1 x 2a 0 b 0, G 2 = a 1b 0 x 2 + a 0 b 1 x 3a 0 b 0The G i ’s are polynomials, therefore the denominator a 0 b 0 divides the numeratorin each of the three expressions on the right hand side. Moreover, if p is a primefactor of a 0 , then p divides the three products b 0 x 0 , b 0 x 1 , b 0 x 2 , hence p divides b 0 .We can repeat the reasoning for a prime divisor of b 0 , so obtaining that a 0 = b 0(up to invertible constants). We get:G 0 = a 1x 0 + b 1 x 1b 0, G 1 = a 1x 1 + b 1 x 2b 0, G 2 = a 1x 2 + b 1 x 3b 0,therefore b 0 divides the numeratorsc 0 := a 1 x 0 + b 1 x 1 , c 1 := a 1 x 1 + b 1 x 2 , c 2 := a 1 x 2 + b 1 x 3 .Hence b 0 divides also x 1 c 0 − x 0 c 1 = b 1 (x 2 1 − x 0x 1 ) = −b 1 F 2 , and similarly x 2 c 0 −x 0 c 2 = b 1 F 1 , x 2 c 1 − x 1 c 2 = −b 1 F 0 . But F 0 , F 1 , F 2 are irreducible and coprime, sowe conclude that b 0 | b 1 . But b 0 and b 1 are coprime, so finally we get b 0 = a 0 = 1.□

Introduction to algebraic geometry 23As a by-product of the proof of Proposition 5.4 we have the minimal freeresolution of the R-module I h (X), where R = K[x 0 , x 1 , . . ., x 3 ]:0 → R ⊕2 ψ−→ R ⊕3 φ−→ I h (X) → 0where ψ is represented by the transposed of the matrix M and φ by the triple ofpolynomials (F 0 , F 1 , F 2 ).Exercises to §5.1*. Let X ⊂ A n be a closed subset, X be its projective closure in P n . Provethat I h (X) = I h (X).2. Find a system of generators of the ideal of the affine skew cubic X, suchthat, if you homogeneize them, you get a system of generators for I h (X).6. Irreducible components.6.1. Definition. Let X ≠ ∅ be a topological space. X is irreducible if thefollowing condition holds: if X 1 , X 2 are closed subsets of X such that X = X 1 ∪X 2 ,then either X = X 1 or X = X 2 . Equivalently, X is irreducible if for all pair of non–empty open subsets U, V we have U ∩ V ≠ ∅. By definition, ∅ is not irreducible.6.2. Proposition. X is irreducible if and only if any non–empty open subset Uof X is dense.Proof. Let X be irreducible, let P be a point of X and I P be an open neighbourhoodof P in X. I P and U are non–empty and open, so I P ∩ U ≠ ∅, therefore P ∈ U.This proves that U = X.Conversely, assume that open subsets are dense. Let U, V ≠ ∅ be opensubsets. Let P ∈ U be a point. By assumption P ∈ V = X, so V ∩ U ≠ ∅ (U isan open neighbourhood of P).□Examples.1. If X = {P } a unique point, then X is irreducible.2. Let K be an infinite field. Then A 1 is irreducible, because proper closedsubsets are finite sets. The same holds for P 1 .3. Let f : X → Y be a continuous map of topological spaces. If X isirreducible and f is surjective, then Y is irreducible.4. Let Y ⊂ X be a subset, give it the induced topology. Then Y is irreducibleif and only if the following holds: if Y ⊂ Z 1 ∪ Z 2 , with Z 1 and Z 2 closed in X,then either Y ⊂ Z 1 or Y ⊂ Z 2 ; equivalently: if Y ∩ U ≠ ∅, Y ∩ V ≠ ∅, with U, Vopen subsets of X, then Y ∩ U ∩ V ≠ ∅.6.3. Proposition. Let X be a topological space, Y a subset of X. Y isirreducible if and only if Y is irreducible.

24 MezzettiProof. Note first that if U ⊂ X is open and U ∩Y = ∅ then U ∩Y = ∅. Otherwise,if P ∈ U ∩Y , let A be an open neighbourhood of P: then A∩Y ≠ ∅. In particular,U is an open neighbourhood of P so U ∩ Y ≠ ∅.Let Y be irreducible. If U and V are open subsets of X such that U ∩Y ≠ ∅,V ∩ Y ≠ ∅, then U ∩ Y ≠ ∅ and V ∩ Y ≠ ∅ so Y ∩ U ∩ V ≠ ∅ by irreducibility ofY . Hence Y ∩ (U ∩ V ) ≠ ∅. So Y is irreducible. If Y is irreducible, we get theirreducibility of Y in a completely analogous way.□6.4. Corollary. Let X be an irreducible topological space and U be a non–emptyopen subset of X. Then U is irreducible.Proof. By Proposition 6.2 U = X which is irreducible. By Proposition 6.3 U isirreducible.□For algebraic sets (both affine and projective) irreducibility can be expressedin a purely algebraic way.6.5. Proposition. Let X ⊂ A n ( resp. P n ) be an algebraic set. X is irreducibleif and only if I(X) (resp. I h (X)) is prime.Proof. Assume first that X is irreducible, X ⊂ A n . Let F, G polynomials ofK[x 1 , . . ., x n ] such that FG ∈ I(X): thenV (F) ∪ V (G) = V (FG) ⊃ V (I(X)) = Xhence either X ⊂ V (F) or X ⊂ V (G). In the former case, if P ∈ X then F(P) = 0,so F ∈ I(X), in the second case G ∈ I(X); hence I(X) is prime.Assume now that I(X) is prime. Let X = X 1 ∪X 2 be the union of two closedsubsets. Then I(X) = I(X 1 ) ∩ I(X 2 ) (see §4). Assume that X 1 ≠ X, then I(X 1 )strictly contains I(X) (otherwise V (I(X 1 )) = V (I(X)). So there exists F ∈ I(X 1 )such that F ∉ I(X). But for every G ∈ I(X 2 ), FG ∈ I(X 1 )∩I(X 2 ) = I(X) prime:since F ∉ I(X), then G ∈ I(X). So I(X 2 ) ⊂ I(X) hence I(X 2 ) = I(X).If X ⊂ P n , the proof is similar, taking into account the following:6.6. Lemma Let P ⊂ K[x 0 , x 1 , . . ., x n ] be a homogeneous ideal. Then P is primeif and only if, for every pair of homogeneous polynomials F, G such that FG ∈ P,either F ∈ P or G ∈ P.Proof of the Lemma. Let H, K be any polynomials such that HK ∈ P. LetH = H 0 + H 1 + . . . + H d , K = K 0 + K 1 + . . . + K e (with H d ≠ 0 ≠ K e ) be theirexpressions as sums of homogeneous polynomials. Then HK = H 0 K 0 + (H 0 K 1 +H 1 K 0 ) + . . . + H d K e : the last product is the homogeneous component of degreed + e of HK. P being homogeneous, H d K e ∈ P; by assumption either H d ∈ P orK e ∈ P. In the former case, HK − H d K = (H − H d )K belongs to P while in thesecond one H(K − K e ) ∈ P. So in both cases we can proceed by induction. □

Introduction to algebraic geometry 25We list now some consequences of the previous Proposition.1. Let K be an infinite field. Then A n and P n are irreducible, becauseI(A n ) = I h (P n ) = (0).2. Let Y ⊂ P n be closed. Y is irreducible if and only if its affine cone C(Y )is irreducible.3. Let Y = V (F) ⊂ A n , be a hypersurface over an algebraically closed fieldK. If F is irreducible, then Y is irreducible.4. Let K be algebraically closed. There is a bijection between prime ideals ofK[x 1 , . . ., x n ] and irreducible algebraic subsets of A n . In particular, the maximalideals correspond to the points. Similarly, there is a bijection between homogeneousnon–irrelevant prime ideals of K[x 0 , x 1 , . . ., x n ] and irreducible algebraicsubsets of P n .6.7. Definition. A topological space X is called noetherian if it satisfies thefollowing equivalent conditions:(i) the ascending chain condition for open subsets;(ii) the descending chain condition for closed subsets;(iii) any non–empty set of open subsets of X has maximal elements;(iv) any non–empty set of closed subsets of X has minimal elements.The proof of the equivalence is standard.Example.subsetsA n is noetherian: if the following is a descending chain of closedY 1 ⊃ Y 2 ⊃ . . . ⊃ Y k ⊃ . . .,thenI(Y 1 ) ⊂ I(Y 2 ) ⊂ . . . ⊂ I(Y k ) ⊂ . . .is an ascending chain of ideals of K[x 1 , . . ., x n ] hence stationary from a suitable mon; therefore V (I(Y m )) = Y m = V (I(Y m )) = Y m+1 = . . ..6.8. Proposition. Let X be a noetherian topological space and Y be a non–empty closed subset of X. Then Y can be written as a finite union Y = Y 1 ∪. . . ∪ Y r of irreducible closed subsets. The maximal Y i ’s in the union are uniquelydetermined by Y and called the “ irreducible components” of Y . They are themaximal irreducible subsets of Y .Proof. By contradiction. Let S be the set of the non–empty closed subsets of Xwhich are not a finite union of irreducible closed subsets: assume S ≠ ∅. Bynoetherianity S has minimal elements, fix one of them Z. Z is not irreducible, soZ = Z 1 ∪ Z 2 , Z i ≠ Z for i = 1, 2. So Z 1 , Z 2 ∉ S, hence Z 1 , Z 2 are both finiteunions of irreducible closed subsets, so such is Z: a contradiction.Now assume that Y = Y 1 ∪ . . . ∪ Y r , with Y i ⊈ Y j if i ≠ j and Y i irreducibleclosed for all i. If there is another similar expression Y = Y 1 ′ ∪ . . . ∪ Y s, ′ Y i ′ ⊈ Y j′

26 Mezzettifor i ≠ j, then Y ′1 ⊂ Y 1 ∪ . . .Y r , so Y ′1 = ⋃ ri=1 (Y ′i, and we can assume i = 1. Similarly, Y 1 ⊂ Y j ′so j = 1 and Y 1 = Y ′1 ∪ Y i ), hence Y 1 ′ ⊂ Y i for some′, for some j, so Y 1 ⊂ Y 1 ⊂ Y ′1. Now let Z = Y − Y 1 = Y 2 ∪ . . . ∪ Y r = Y 2 ′ ∪ . . . ∪ Y s ′ andproceed by induction.□j ,6.9. Corollary. Any algebraic subset of A n (resp. of P n ) is in a unique way thefinite union of its irreducible components.□Note that the irreducible components of X are its maximal algebraic subsets.They correspond to the minimal prime ideals over I(X). Since I(X) is radical,these minimal prime ideals coincide with the primary ideals appearing in the primarydecomposition of I(X).6.10. Definition. An irreducible closed subset of A n is called an affine variety.Similarly, an irreducible closed subset of P n is a projective variety. A locally closedsubset in P n is the intersection of an open and a closed subset. An irreduciblelocally closed subset of P n is a quasi–projective variety.6.11. Proposition. Let X ⊂ A n and Y ⊂ A m be affine varieties. Then X × Yis irreducible, i.e. a subvariety of A n+m .Proof. Let X × Y = W 1 ∪ W 2 , with W 1 , W 2 closed. For all P ∈ X the map{P } × Y → Y which takes (P, Q) to Q is a homeomorphism, so {P } × Y isirreducible. {P } × Y = (W 1 ∩ ({P } × Y )) ∪ (W 2 ∩ ({P } × Y )), so ∃i ∈ {1, 2} suchthat {P } × Y ⊂ W i . Let X i = {P ∈ X | {P } × Y ⊂ W i }, i = 1, 2. Note thatX = X 1 ∪ X 2 .Claim. X i is closed in X.Let X i (Q) = {P ∈ X | (P, Q) ∈ W i }, Q ∈ Y . We have: (X × {Q}) ∩ W i =X i (Q) × {Q} ≃ X i (Q); X × {Q} and W i are closed in X × Y , so X i (Q) × {Q}is closed in X × Y and also in X × {Q}, so X i (Q) is closed in X. Note thatX i = ⋂ Q∈Y Xi (Q), hence X i is closed, which proves the Claim.Since X is irreducible, X = X 1 ∪ X 2 implies that either X = X 1 or X = X 2 ,so either X × Y = W 1 or X × Y = W 2 .□Exercises to §6.1. Let X ≠ ∅ be a topological space. Prove that X is irreducible if and onlyif all non–empty open subsets of X are connected.2*. Prove that the cuspidal cubic Y ⊂ A 2 C of equation x3 −y 2 = 0 is irreducible.(Hint: express Y as image of A 1 in a continuous map...)

Introduction to algebraic geometry 273. Give an example of two irreducible subvarieties of P 3 whose intersection isreducible.4. Find the irreducible components of the following algebraic sets over thecomplex field:a) V (y 4 − x 2 , y 4 − x 2 y 2 + xy 2 − x 3 ) ⊂ A 2 ;b) V (y 2 − xz, z 2 − y 3 ) ⊂ A 3 .5*. Let Z be a topological space and {U α } α∈I be an open covering of Z suchthat U α ∩ U β ≠ ∅ for α ≠ β and that all U α ’s are irreducible. Prove that Z isirreducible.7. Dimension.Let X be a topological space.7.1. Definition. The topological dimension of X is the supremum of the lengthsof the chains of distinct irreducible closed subsets of X, where by definiton thefollowing chain has length n:X 0 ⊂ X 1 ⊂ X 2 ⊂ . . . ⊂ X n .Example.1. dim A 1 = 1: the maximal length chains have the form {P } ⊂ A 1 .2. dim A n = n: a chain of length n is{0} = V (x 1 , . . ., x n ) ⊂ V (x 1 , . . ., x n−1 ) ⊂ . . . ⊂ V (x 1 ) ⊂ A n ;note that V (x 1 , . . ., x i ) is irreducible ∀i, because the ideal 〈x 1 , . . ., x i 〉 is prime.Indeed the quotient K[x 1 , . . ., x n ]/〈x 1 , . . ., x i 〉 is isomorphic to K[x i+1 , . . ., x n ].Therefore dim A n ≥ n. On the other hand, from every chain of irreducible closedsubsets of A n , passing to the ideals, we get a chain of the same length of primeideals in K[x 1 , . . ., x n ], therefore dim A n ≤ dim K[x 1 , . . ., x n ] = n.3. Let X be irreducible. Then dim X = 0 if and only if X is the closure ofevery point of it.We prove now some useful relations between the dimension of X and thedimensions of its subspaces.7.2. Proposition.1. If Y ⊂ X, then dim Y ≤ dimX. In particular, if dimX is finite, then alsodim Y is (in this case, the number dim X − dimY is called the codimension of Yin X).2. If X = ⋃ i∈I U i is an open covering, then dim X = sup{dim U i }.3. If X is noetherian and X 1 , . . ., X s are its irreducible components, thendim X = sup i dim X i .

28 Mezzetti4. If Y ⊂ X is closed, X is irreducible, dim X is finite and dimX = dim Y ,then Y = X.Proof.1. Let Y 0 ⊂ Y 1 ⊂ . . . ⊂ Y n be a chain of irreducible closed subsets of Y . Thentheir closures are irreducible and form the following chain: Y 0 ⊆ Y 1 ⊆ . . . ⊆ Y n .Note that for all i Y i ∩ Y = Y i , because Y i is closed into Y , so if Y i = Y i+1 , thenY i = Y i+1 . Therefore the two chains have the same length and we can concludethat dim Y ≤ dimX.2. Let X 0 ⊂ X 1 ⊂ . . . ⊂ X n be a chain of irreducible closed subsets of X. LetP ∈ X 0 be a point: there exists an index i ∈ I such that P ∈ U i . So ∀k = 0, . . ., nX k ∩ U i ≠ ∅: it is an irreducible closed subset of U i , irreducible because open inX k which is irreducible. Consider X 0 ∩U i ⊂ X 1 ∩U i ⊂ . . . ⊂ X n ∩U i ; it is a chainof length n, because X k ∩ U i = X k : in fact X k ∩ U i is open in X k hence dense.Therefore, for all chain of irreducible closed subsets of X, there exists a chain ofthe same length of irreducible closed subsets of some U i . So dim X ≤ sup dim U i .By 1., equality holds.3. Any chain of irreducible closed subsets of X is completely contained in anirreducible component of X. The conclusion follows as in 2.4. If Y 0 ⊂ Y 1 ⊂ . . . ⊂ Y n is a chain of maximal length in Y , then it is amaximal chain in X, because dim X = dim Y . Hence X = Y n ⊂ Y . □7.3. Corollary. dim P n = dim A n .Proof. Because P n = U 0 ∪ . . . ∪ U n , and U i is homeomorphic to A n for all i.□If X is noetherian and all its irreducible components have the same dimensionr, then X is said to have pure dimension r.Note that the topological dimension is invariant by homeomorphism. Bydefinition, a curve is an algebraic set of pure dimension 1; a surface is an algebraicset of pure dimension 2.We want to study the dimensions of affine algebraic sets. The following definitionresults to be very important.7.4. Definition. Let X ⊂ A n be an algebraic set. The coordinate ring of X isK[X] := K[x 1 , . . ., x n ]/I(X).It is a finitely generated K–algebra without non–zero nilpotents, because I(X)is radical. There is the canonical epimorphism K[x 1 , . . ., x n ] → K[X] such thatF → [F]. The elements of K[X] can be interpreted as polynomial functions onX: to a polynomial F, we can associate the function f : X → K such thatP(a 1 , . . ., a n ) → F(a 1 , . . ., a n ).Two polynomials F, G define the same function on X if, and only if, F(P) =G(P) for every point P ∈ X, i.e. if F − G ∈ I(X), which means exactly that Fand G have the same image in K[X].

Introduction to algebraic geometry 29K[X] is generated as K–algebra by [x 1 ], . . ., [x n ]: these can be interpreted asthe functions on X called coordinate functions, and generally denoted t 1 , . . ., t n .In fact t i : X → K is the function which associates to P(a 1 , . . ., a n ) the constanta i . Note that the function f can be interpreted as F(t 1 , . . ., t n ): the polynomialF evalued at the n– tuple of the coordinate functions.In the projective space we can do an analogous construction. If Y ⊂ P n isclosed, then the homogeneous coordinate ring of Y isS(Y ) := K[x 0 , x 1 , . . ., x n ]/I h (Y ).It is also a finitely generated K–algebra, but its elements have no interpretationas functions on Y . They are functions on the cone C(Y ).7.5. Definition. Let R be a ring. The Krull dimension of R is the supremumof the lengths of the chains of prime ideals of RP 0 ⊂ P 1 ⊂ . . . ⊂ P r .Similarly, the heigth of a prime ideal P is the sup of the lengths of the chains ofprime ideals contained in P: it is denoted htP.7.6. Proposition. Let K be an algebraically closed field. Let X be an affinealgebraic set contained in A n . Then dimX = dim K[X].Proof.By the Nullstellensatz and by 6.5 the chains of irreducible closed subsets ofX correspond bijectively to the chains of prime ideals of K[x 1 , . . ., x n ] containingI(X), hence to the chains of prime ideals of the quotient ring K[X]. □The dimension theory for commutative rings contains some important theoremsabout dimension of K–algebras. The following two results are very useful.7.7. Theorem. Let K be any field.1. Let B be a finitely generated K–algebra and an integral domain. Thendim B = tr.d.Q(B)/K, where Q(B) is the quotient field of B. In particular dimBis finite.2. Let B be as above and P ⊂ B be any prime ideal. Then dim B =htP + dim B/P.Proof. For 1. see Portelli’s notes. For a proof of 2., see for instance [4], Ch. II,Proposition 3.4. It relies on the normalization lemma and the lying over theorem.□7.8. Corollary. Let K be an algebraically closed field.1. dim A n = dim P n = n.

30 Mezzetti2. If X is an affine variety, then dim X = tr.d.K(X)/K, where K(X) denotesthe quotient field of K[X].2. If X ⊂ A n is closed and irreducible, then dim X = n − htI(X). □The following is an important characterization of the algebraic subsets of A nof codimension 1.7.9. Proposition. Let X ⊂ A n be closed. Then X is a hypersurface if and onlyif X is of pure dimension n − 1.Proof. We give here an elementary direct proof. It can be proved more quicklyusing the Krull principal ideal theorem.Let X ⊂ A n be a hypersurface, with I(X) = (F) = (F r 11 . . .Fr ss ), whereF 1 , . . ., F s are the irreducible factors of F with multiplicities r 1 , . . ., r s respectively.Then V (F 1 ),. . ., V (F s ) are the irreducible components of X, whose ideals are (F 1 ),. . ., (F s ). So it is enough to prove that ht(F i ) = 1, for i = 1, . . ., s.If P ⊂ (F i ) is a prime ideal, then either P = (0) or there exists G ∈ P, G ≠ 0.In the second case, let A be an irreducible factor of G belonging to P: A ∈ (F i )so A = HF i . Since A is irreducible, either H or F i is invertible; F i is irreducible,so H is invertible, hence (A) = (F i ) ⊂ P. Therefore either P = (0) or P = (F i ),and ht(F i ) = 1.Conversely, assume that X is irreducible of dimension n − 1. Since X ≠ A n ,there exists F ∈ I(X), F ≠ 0. By the irreducibility, some irreducible factor ofF, call it H, also vanishes along X. Therefore X ⊂ V (H), which is irreducible ofdimension n − 1, by the first part. So X = V (H) (by Proposition 7.2, 3). □This proposition does not generalize to higher codimension. There exist codimension2 algebraic subsets of A n whose ideal is not generated by two polynomials.An example in A 3 is the curve X parametrized by (t 3 , t 4 , t 5 ). A system of generatorsof I(X) is 〈x 3 − yz, y 2 − xz, z 2 − x 2 y〉. One can show that I(X) cannot begenerated by two polynomials. For a discussion of this and other similar examples,see [4], Chapter V.7.10. Proposition. Let X ⊂ A n , Y ⊂ A m be irreducible closed subsets. Thendim X × Y = dim X + dimY .Proof. Let r = dim X, s = dim Y ; let t 1 , . . ., t n (resp. u 1 , . . ., u m ) be coordinatefunctions on A n (resp. A m ). We can assume that t 1 , . . ., t r be a transcendence basisof Q(K[X]) and u 1 , . . ., u s be a transcendence basis of Q(K[Y ]). By definition,K[X × Y ] is generated as K–algebra by t 1 , . . ., t n , u 1 , . . ., u m : we want to showthat t 1 , . . ., t r , u 1 , . . ., u s is a transcendence basis of Q(K[X ×Y ]) over K. Assumethat F(x 1 , . . ., x r , y 1 , . . ., y s ) is a polynomial which vanishes on t 1 , . . ., t r , u 1 , . . ., u s ,i.e. F defines the zero function on X × Y . Then, ∀ P ∈ X, F(P; y 1 , . . ., y s )is zero on Y , i.e. F(P; u 1 , . . ., u s ) = 0. Since u 1 , . . ., u s are algebraically independent,every coefficient a i (P) of F(P; y 1 , . . ., y s ) is zero, ∀ P ∈ X. Since

Introduction to algebraic geometry 31t 1 , . . ., t r are algebraically independent, the polynomials a i (x 1 , . . ., x r ) are zero, soF(x 1 , . . ., x r , y 1 , . . ., y s ) = 0. So t 1 , . . ., t r , u 1 , . . ., u s are algebraically independent.Since this is certainly a maximal algebraically free set, it is a transcendence basis.□Exercises to §7.1*. Prove that a proper closed subset of an irreducible curve is a finite set.Deduce that any bijection between irreducible curves is a homeomorphism.2*. Let X ⊂ A 2 be the cuspidal cubic of equation: x 3 − y 2 = 0, let K[X] beits coordinate ring. Prove that all elements of K[X] can be written in a uniqueway in the form f(x)+yg(x), where f, g are polynomial in the variable x. Deducethat K[X] is not isomorphic to a polynomial ring.8. Regular and rational functions.Let X ⊂ P n be a locally closed subset and P be a point of X. Let φ : X → K bea function.8.1. Definition. φ is regular at P if there exists a suitable neighbourhood ofP in which it can be expressed as a quotient of homogeneous polynomials of thesame degree; more precisely, if there exist an open neighbourhood U of P in X andhomogeneous polynomials F, G ∈ K[x 0 , x 1 , . . ., x n ] with deg F = deg G, such thatU ∩ V P (G) = ∅ and φ(Q) = F(Q)/G(Q), for all Q ∈ U. Note that the quotientF(Q)/G(Q) is well defined.φ is regular on X if φ is regular at every point P of X.The set of regular functions on X is denoted O(X): it contains K (identified withthe set of constant functions), and can be given the structure of a K–algebra, bythe definitions:(φ + ψ)(P) = φ(P) + ψ(P)(φψ)(P) = φ(P)ψ(P),for P ∈ X. (Check that φ + ψ and φψ are indeed regular on X.)8.2. Proposition. Let φ : X → K be a regular function. Let K be identifiedwith A 1 with Zariski topology. Then φ is continuous.Proof. It is enough to prove that φ −1 (c) is closed in X, ∀ c ∈ K. For all P ∈ X,choose an open neighbourhood U P and homogeneous polynomials F P , G P suchthat φ| P = F P /G P . Thenφ −1 (c) ∩ U P = {Q ∈ U P |F P (Q) − cG P (Q) = 0} = U P ∩ V P (F P − cG P )

32 Mezzettiis closed in U P . The proposition then follows from:8.3. Lemma. Let T be a topological space, T = ∪ i∈I U i be an open covering ofT, Z ⊂ T be a subset. Then Z is closed if and only if Z ∩ U i is closed in U i for alli.Proof. Assume that U i = X \ C i and Z ∩ U i = Z i ∩ U i , with C i and Z i closed inX.Claim: Z = ⋂ i∈I (Z i ∪ C i ), hence it is closed.In fact: if P ∈ Z, then P ∈ Z ∩ U i for a suitable i. Therefore P ∈ Z i ∩ U i , soP ∈ Z i ∪ C i . If P /∈ Z j ∩ U j for some j, then P /∈ U j so P ∈ C j and thereforeP ∈ Z j ∪ C j .Conversely, if P ∈ ⋂ i∈I (Z i ∪ C i ), then ∀ i, either P ∈ Z i or P ∈ C i . Since ∃jsuch that P ∈ U j , hence P /∈ C j , so P ∈ Z j , so P ∈ Z j ∩ U j = Z ∩ U j .□8.4. Corollary.1. Let φ ∈ O(X): then φ −1 (0) is closed. It is denoted V (φ) and called theset of zeroes of φ.2. Let X be a quasi–projective variety and φ, ψ ∈ O(X). Assume that thereexists U, open non –empty subset such that φ| U = ψ| U . Then φ = ψ.Proof. φ − ψ ∈ O(X) so V (φ − ψ) is closed. By assumption V (φ − ψ) ⊃ U, whichis dense, because X is irreducible. So V (φ − ψ) = X.□If X ⊂ A n is locally closed, we can use on X both homogeneous and non–homogeneous coordinates. In the second case, a regular function is locally representedas a quotient F/G, with F and G ∈ K[x 1 , . . ., x n ]. In particular allpolynomial functions are regular, so, if X is closed, K[X] ⊂ O(X).If α ⊂ K[X] is an ideal, we can consider V (α) := ⋂ φ∈αV (φ): it is closedinto X. Note that α is of the form α = α/I(X), where α is the image of α inthe canonical epimorphism, it is an ideal of K[x 1 , . . ., x n ] containing I(X), henceV (α) = V (α) ∩ X = V (α).If K is algebraically closed, from the Nullstellensatz it follows that, if α isproper, then V (α) ≠ ∅. Moreover the following relative form of the Nullstellensatzholds: if f ∈ K[X] and f vanishes at all points P ∈ X such that g 1 (P) = . . . =g m (P) = 0 (g 1 , . . ., g m ∈ K[X]), then f r ∈ 〈g 1 , . . ., g m 〉 ⊂ K[X], for some r ≥ 1.8.5. Theorem. Let K be an algebraically closed field. Let X ⊂ A n K be closedin the Zariski topology. Then O(X) ≃ K[X]. It is an integral domain if and onlyif X is irreducible.Proof. Let f ∈ O(X).

Introduction to algebraic geometry 33(i) Assume first that X is irreducible. For all P ∈ X fix an open neighbourhoodU P of P and polynomials F P , G P such that V P (G P ) ∩ U P = ∅ andf| UP = F P /G P . Let f P , g P be the functions of K[X] defined by F P and G P . Theng P f = f P holds on U P , so it holds on X (by Corollary 8.3). Let α ⊂ K[X] bethe ideal α = 〈g P 〉 P ∈X ; α has no zeroes on X, because g P (P) ≠ 0, so α = K[X].Therefore there exists h P ∈ K[X] such that 1 = ∑ P ∈X h Pg P (sum with finitesupport). Hence in O(X) we have the relation: f = f ∑ h P g P = ∑ h P (g P f) =∑hP f P ∈ K[X].(ii) Let X be reducible: for all P ∈ X, there exists R ∈ K[x 1 , . . ., x n ] such thatR(P) ≠ 0 and R ∈ V (X \ U P ), so r ∈ O(X) is zero outside U P . So rg P f = f P ron X and we conclude as above by replacing g P with g P r and f P with f P r.□The characterization of regular functions on projective varieties is completely different:we will see in §12 that, if X is a projective variety, then O(X) ≃ K, i.e.the unique regular functions are constant.This gives the motivation for introducing the following weaker concept.8.6. Definition. Let X be a quasi–projective variety. A rational function on Xis a germ of regular functions on some open non–empty subset of X.Precisely, let K be the set {(U, f)|U ≠ ∅, open subset of X, f ∈ O(U)}. Thefollowing relation on K is an equivalence relation:(U, f) ∼ (U ′ , f ′ ) if and only if f| U∩U′ = f ′ | U∩U ′.Reflexive and symmetric properties are quite obvious. Transitive property: let(U, f) ∼ (U ′ , f ′ ) and (U ′ , f ′ ) ∼ (U ′′ , f ′′ ). Then f| U∩U′ = f ′ | U∩U′ and f ′ | U ′ ∩U ′′ =f ′′ | U ′ ∩U ′′, hence f| U∩U ′ ∩U ′′ = f ′′ | U∩U ′ ∩U ′′. U ∩ U ′ ∩ U ′′ is a non–empty opensubset of U ∩ U ′′ (which is irreducible and quasi–projective), so by Corollary 8.4f| U ′ ∩U ′′ = f ′′ | U ′ ∩U ′′.Let K(X) := K/ ∼: its elements are by definition rational functions on X.K(X) can be given the structure of a field in the following natural way.Let 〈U, f〉 denote the class of (U, f) in K(X). We define:〈U, f〉 + 〈U ′ , f ′ 〉 = 〈U ∩ U ′ , f + f ′ 〉,〈U, f〉〈U ′ , f ′ 〉 = 〈U ∩ U ′ , ff ′ 〉(check that the definitions are well posed!).There is a natural inclusion: K → K(X) such that c → 〈X, c〉. Moreover, if〈U, f〉 ≠ 0, then there exists 〈U, f〉 −1 = 〈U \ V (f), f −1 〉: the axioms of a field areall satisfied.There is also an injective map: O(X) → K(X) such that φ → 〈X, φ〉.

34 Mezzetti8.7. Proposition. If X ⊂ A n is affine, then K(X) ≃ Q(O(X)) = K(t 1 , . . ., t n ),where t 1 , . . ., t n are the coordinate functions on X.Proof. The isomorphism is as follows:(i) ψ : K(X) → Q(O(X))If 〈U, φ〉 ∈ K(X), then there exists V ⊂ U, open and non–empty, such that φ | V =F/G, where F, G ∈ K[x 1 , . . ., x n ] and V (G) ∩ V = ∅. We set ψ(〈U, φ〉) = f/g.(ii) ψ ′ : Q(O(X)) → K(X)If f/g ∈ Q(O(X)), we set ψ ′ (f/g) = 〈X \ V (g), f/g〉.It is easy to check that ψ and ψ ′ are well defined and inverse each other. □8.8. Corollary. If X is an affine variety, then dimX is equal to the transcendencedegree over K of its field of rational functions..8.9. Proposition. If X is quasi–projective and U ≠ ∅ is an open subset, thenK(X) ≃ K(U).Proof. We have the maps: K(U) → K(X) such that 〈V, φ〉 → 〈V, φ〉, and K(X) →K(U) such that 〈A, ψ〉 → 〈A ∩ U, ψ | A∩U 〉: they are K–homomorphisms inverseeach other.□8.10. Corollary. If X is a projective variety contained in P n , if i is an indexsuch that X ∩ U i ≠ ∅, then dimX = dimX ∩ U i = tr.d.K(X)/K.Proof. By Proposition 7.2 dimX = sup dim(X ∩ U i ). By 8.8 and 8.9, if X ∩ U i isnon–empty, dim(X ∩ U i ) = tr.d.K(X ∩ U i )/K = tr.d.K(X)/K is independent ofi. □If 〈U, φ〉 ∈ K(X), we can consider all possible representatives of it, i.e. allpairs 〈U i , φ i 〉 such that 〈U, φ〉 = 〈U i , φ i 〉. Then U = ⋃ i U i is the maximum opensubset of X on which φ can be seen as a function: it is called the domain ofdefinition (or of regularity) of 〈U, φ〉, or simply of φ. It is sometimes denoteddomφ. If P ∈ U, we say that φ is regular at P.We can consider the set of rational functions on X which are regular at P: itis denoted O P,X . It is a subring of K(X) containing O(X), called the local ringof X at P. In fact, O P,X is a local ring, whose maximal ideal, denoted M P,X , isformed by rational functions φ such that φ(P) is defined and φ(P) = 0. To seethis, observe that an element of O P,X can be represented as 〈U, F/G〉: its inversein K(X) is 〈U \ V P (G), G/F 〉, which belongs to O P,X if and only if F(P) ≠ 0.We’ll see in 8.12 that O P,X is the localization K[X] IX (P).As in Proposition 8.9 for the fields of rational functions, also for the local ringsof points it can easily be proved that, if U ≠ ∅ is an open subset of X containingP, then O P,X ≃ O P,U . So the ring O P,X only depends on the local behaviour of

Introduction to algebraic geometry 35X in the neighbourhood of P.The residue field of O P,X is the quotient O P,X /M P,X : it is a field which resultsto be naturally isomorphic to the base field K. In fact consider the evaluationmap O P,X → K such that φ goes to φ(P): it is surjective with kernel M P,X , soO P,X /M P,X ≃ K.8.11. Examples.1. Let Y ⊂ A 2 be the curve V (x 3 1 − x2 2 ). Then F = x 2, G = x 1 define thefunction φ = x 2 /x 1 which is regular at the points P(a 1 , a 2 ) such that a 1 ≠ 0.Another representation of the same function is: φ = x 2 1 /x 2, which shows thatφ is regular at P if a 2 ≠ 0. If φ admits another representation F ′ /G ′ , thenG ′ x 2 − F ′ x 1 vanishes on an open subset of X, which is irreducible (see Exercise6.2), hence G ′ x 2 − F ′ x 1 vanishes on X, and therefore G ′ x 2 − F ′ x 1 ∈ 〈x 3 1 − x2 2 〉.This shows that there are essentially only the above two representations of φ. Soφ ∈ K(X) and its domain of regularity is Y \ {0, 0}.2. The stereographic projection.Let X ⊂ P 2 be the curve V P (x 2 1 +x2 2 −x2 0 ). Let f := x 1/(x 0 −x 2 ) denote the germ ofthe regular function defined by x 1 /(x 0 −x 2 ) on X \V P (x 0 −x 2 ) = X \ {[1, 0, 1]} =X \ {P }. On X we have x 2 1 = (x 0 − x 2 )(x 0 + x 2 ) so f is represented also as(x 0 + x 2 )/x 1 on X \ V P (x 1 ) = X \ {P, Q}, where Q = [1, 0, −1]. If we identify Kwith the affine line V P (x 2 ) \ V P (x 0 ) (the points of the x 1 –axis lying in the affineplane U 0 ), then f can be interpreted as the stereographic projection of X centeredat P, which takes A[a 0 , a 1 , a 2 ] to the intersection of the line AP with the lineV P (x 2 ). To see this, observe that AP has equation a 1 x 0 +(a 2 − a 0 )x 1 −a 1 x 2 = 0;and AP ∩ V P (x 2 ) is the point [a 0 − a 2 , a 1 , 0].8.12. The algebraic characterization of the local ring O P,X .Let us recall the construction of the ring of fractions of a ring A with respectto a multiplicative subset S.Let A be a ring and S ⊂ A be a multiplicative subset. The following relationin A × S is an equivalence relation:(a, s) ≃ (b, t) if and only if ∃u ∈ S such that u(at − bs) = 0.Then the quotient A × S/ ≃ is denoted S −1 A or A S and [(a, s)] is denoted a s . A Sbecomes a commutative ring with unit with operations a s + b t = at+bsstand a bs t = abst(check that they are well–defined). With these operations, A S is called the ring offractions of A with respect to S, or the localization of A in S.There is a natural homomorphism j : A → S −1 A such that j(a) = a 1 , whichmakes S −1 A an A–algebra. Note that j is the zero map if and only if 0 ∈ S. Moreprecisely if 0 ∈ S then S −1 A is the zero ring: this case will always be excluded inwhat follows. Moreover j is injective if and only if S is formed by regular elements.In this case j(A) will be identified with A.

36 MezzettiExamples.1. Let A be an integral domain and set S = A \ {0}. Then A S = Q(A): thequotient field of A.2. If P ⊂ A is a prime ideal, then S = A \ P is a multiplicative set and A S isdenoted A P and called the localization of A at P.3. If f ∈ A, then the multiplicative set generated by f isS = {1, f, f 2 , . . ., f n , . . .} :A S is denoted A f .4. If S = {x ∈ A | x is regular}, then A S is called the total ring of fractionsof A: it is the maximum ring in which A can be canonically embedded.It is easy to verify that the ring A S enjoys the following universal property:(i) if s ∈ S, then j(s) is invertible;(ii) if B is a ring with a given homomorphism f : A → B such that if s ∈ S,then f(s) is invertible, then f factorizes through A S , i.e. there exists a uniquehomomorphism f such that f ◦ j = f.We will see now the relations between ideals of A S and ideals of A.If α ⊂ A is an ideal, then αA S = { a s| a ∈ α} is called the extension of α inA S and denoted also α e . It is an ideal, precisely the ideal generated by the set{ a 1 | a ∈ α}.If β ⊂ A S is an ideal, then j −1 (β) =: β c is called the contraction of β and isclearly an ideal.We have:8.13. Proposition.1. ∀α ⊂ A : α ec ⊃ α;2. ∀β ⊂ A S : β = β ce ;3. α e is proper if and only if α ∩ S = ∅;4. α ec = {x ∈ A | ∃s ∈ S such that sx ∈ α}.Proof.1. and 2. are straightforward.3. if 1 = a s ∈ αe , then there exists u ∈ S such that u(s − a) = 0, i.e.us = ua ∈ S ∩ α. Conversely, if s ∈ S ∩ α then 1 = s s ∈ αe .4.α ec = {x ∈ A | j(x) = x 1 ∈ αe } == {x ∈ A | ∃a ∈ α, t ∈ S such that x 1 = a t } == {x ∈ A | ∃a ∈ α, t, u ∈ S such that u(xt − a) = 0}.

Introduction to algebraic geometry 37Hence, if x ∈ α ec , then: (ut)x = ua ∈ α. Conversely: if there exists s ∈ S suchthat sx = a ∈ α, then x 1 = a s , i.e. j(x) ∈ αe .□If α is an ideal of A such that α = α ec , α is called saturated with S. Forexample, if P is a prime ideal and S ∩ P = ∅, then P is saturated and P e is prime.Conversely, if Q ⊂ A S is a prime ideal, then Q c is prime in A.Therefore: there is a bijection between the set of prime ideals of A S and theset of prime ideals of A not intersecting S. In particular, if S = A \ P, P prime,the prime ideals of A P correspond bijectively to the prime ideals of A containedin P, hence A P is a local ring with maximal ideal P e , denoted PA P , and residuefield A P /PA P . Moreover dim A P = htP.In particular we get the characterization of O P,X . Let X ⊂ A n be an affinevariety, let P be a point of X and I(P) ⊂ K[x 1 , . . ., x n ] be the ideal of P. LetI X (P) := I(P)/I(X) be the ideal of K[X] formed by regular functions on Xvanishing at P. Then we can construct the localizationO(X) IX (P) = { f |f, g ∈ O(X), g(P) ≠ 0} ⊂ K(X) :git is canonically identified with O P,X . In particular: dim O P,X = ht I X (P) =dim O(X) = dim X.There is a bijection between prime ideals of O P,X and prime ideals of O(X)contained in I X (P); they also correspond to prime ideals of K[x 1 , . . ., x n ] containedin I(P) and containing I(X).If X is affine, it is possible to define the local ring O P,X also if X is reducible,simply as localization of K[X] at the maximal ideal I X (P). The natural map jfrom K[X] to O P,X is injective if and only if K[X] \ I X (P) does not contain anyzero divisor. A non-zero function f is a zero divisor if there exists a non-zero gsuch that fg = 0, i.e. X = V (f) ∪ V (g) is an expression of X as union of properclosed subsets. For j to be injective it is required that every zero divisor f belongsto I X (P), which means that all the irreducible components of X pass through P.Exercises to §8.1. Prove that the affine varieties and the open subsets of affine varieties arequasi–projective.2. Let X = {P, Q} be the union of two points in an affine space over K.Prove that O(X) is isomorphic to K × K.9. Regular and rational maps.Let X, Y be quasi–projective varieties (or more generally locally closed sets). Letφ : X → Y be a map.

38 Mezzetti9.1. Definition. φ is a regular map or a morphism if(i) φ is continuous;(ii) φ preserves regular functions, i.e. for all U ⊂ Y (U open and non–empty) andfor all f ∈ O(U), then f ◦ φ ∈ O(φ −1 (U)):X↑φ −1 (U)φ−→φ|−→Y↑Uf→KNote that:a) for all X the identity map 1 X : X → X is regular;b) for all X, Y , Z and regular maps X φ → Y , Y ψ → Z, the composite map ψ ◦ φ isregular.An isomorphism of varieties is a regular map which possesses regular inverse,i.e. a regular φ : X → Y such that there exists a regular ψ : Y → X verifyingthe conditions ψ ◦ φ = 1 X and φ ◦ ψ = 1 Y . In this case X and Y are said to beisomorphic, and we write: X ≃ Y .If φ : X → Y is regular, there is a natural K–homomorphism φ ∗ : O(Y ) →O(X), called the comorphism associated to φ, defined by: f → φ ∗ (f) := f ◦ φ.The construction of the comorphism is functorial, which means that:a) 1 ∗ X = 1 O(X);b) (ψ ◦ φ) ∗ = φ ∗ ◦ ψ ∗ .This implies that, if X ≃ Y , then O(X) ≃ O(Y ). In fact, if φ : X → Y is anisomorphism and ψ is its inverse, then φ ◦ψ = 1 Y , so (φ ◦ψ) ∗ = ψ ∗ ◦φ ∗ = (1 Y ) ∗ =1 O(Y ) and similarly ψ ◦ φ = 1 X implies φ ∗ ◦ ψ ∗ = 1 O(X) .9.2. Examples.1) The homeomorphism φ i : U i → A n of Proposition 3.2 is an isomorphism.2) There exist homeomorphisms which are not isomorphisms. Let Y = V (x 3 −y 2 ) ⊂ A 2 . We have seen (see Exercise 7.2) that K[X] ≄ K[A 1 ], hence Y is notisomorphic to the affine line. Nevertheless, the following map is regular, bijectiveand also a homeomorphism (see Exercise 7.1):φ : A 1 → Y such that t → (t 2 , t 3 );{ yφ −1 : Y → A 1 is defined by (x, y) → x if x ≠ 00 if (x, y) = (0, 0).9.3. Proposition. Let φ : X → Y ⊂ A n be a map. Then φ is regular if and onlyif φ i := t i ◦ φ is a regular function on X, for all i = 1, . . ., n (where t 1 , . . ., t n arethe coordinate functions on Y ).Proof. If φ is regular, then φ i = φ ∗ (t i ) is regular by definition.

Introduction to algebraic geometry 39Conversely: if F ∈ K[x 1 , . . ., x n ], thenφ −1 (V (F) ∩ Y ) = {P ∈ X|F(φ(P)) = F(φ 1 , . . ., φ n )(P) = 0} = V (F(φ 1 , . . ., φ n ))(note that F(φ 1 , . . ., φ n ) ∈ O(X): it is the composition of F with the regular functionsφ 1 , . . ., φ n ). In particular φ −1 (V (F)∩Y ) is closed, so we can conclude that φis continuous. If U ⊂ Y and f ∈ O(U), for all points P of U choose open neighbourhoodsU P such that f = F P /G P on U P . So f ◦φ = F P (φ 1 , . . ., φ n )/G P (φ 1 , . . ., φ n )on φ −1 (U P ), hence it is regular on each φ −1 (U P ) and by consequence on φ −1 (U).□If φ : X → Y is a regular map and Y ⊂ A n , by Proposition 9.2. we can representφ in the form φ = (φ 1 , . . ., φ n ), where φ 1 , . . ., φ n ∈ O(X) and φ i = φ ∗ (t i ).If Y is closed in A n , let us recall that t 1 , . . ., t n generate O(Y ), hence φ 1 , . . ., φ ngenerate φ ∗ (O(Y )). This observation is the key for the following important result.9.4. Theorem. Let X be a locally closed set and Y be an affine algebraic set. LetHom(X, Y ) denote the set of regular maps from X to Y and Hom(O(Y ), O(X))denote the set of K– homomorphisms from O(Y ) to O(X).Then the map Hom(X, Y ) → Hom(O(Y ), O(X)), such that φ : X → Y goesto φ ∗ : O(Y ) → O(X), is bijective.Proof. Let Y ⊂ A n and let t 1 , . . ., t n be the coordinate functions on Y , so O(Y ) =K[t 1 , . . ., t n ]. Let u : O(Y ) → O(Y ) be a K–homomorphism: we want to define amorphism u ♯ : X → Y whose associated comorphism is u. By the remark above,if u ♯ exists, its components have to be u(t 1 ), . . ., u(t n ). So we defineu ♯ : X → A nP → (u(t 1 )(P)), . . ., u(t n )(P)).This is a morphism by Proposition 9.3 and we claim that u ♯ (X) ⊂ Y . Let F ∈ I(Y )and P ∈ X: then(F(u ♯ (P)) = F(u(t 1 )(P), . . ., u(t n )(P)) == F(u(t 1 ), . . ., u(t n ))(P) == u(F((t 1 , . . ., t n ))(P) because u is K-homomorphism == u(0)(P) == 0(P) = 0.So u ♯ is a regular map from X to Y .We consider now (u ♯ ) ∗ : O(Y ) → O(X): it takes a function f to f ◦ u ♯ =f(u(t 1 ), . . ., u(t n )) = u(f), so (u ♯ ) ∗ = u. Conversely, if φ : X → Y is regular, then(φ ∗ ) ♯ takes P to (φ ∗ (t 1 )(P), . . ., φ ∗ (t n )(P)) = (φ 1 (P), . . ., φ n (P)), so (φ ∗ ) ♯ = φ.□

40 MezzettiNote that, by definition, 1 ♯ O(X) = 1 X, for all affine X; moreover (v◦u) ♯ = u ♯ ◦v ♯for all u : O(Z) → O(Y ), v : O(Y ) → O(X), K–homomorphisms of affine algebraicsets: this means that also this construction is functorial.9.5. Corollary. Let X, Y be affine algebraic sets. Then X ≃ Y if and only ifO(X) ≃ O(Y ).□If X and Y are quasi–projective varieties and φ : X → Y is regular, it is notalways possible to define a comorphism K(Y ) → K(X). If f is a rational functionon Y with domf = U, it can happen that φ(X) ∩ domf = ∅, in which case f ◦ φdoes not exist. Nevertheless, if we assume that φ is dominant, i.e. φ(X) = Y ,then certainly φ(X) ∩ U ≠ ∅, hence 〈φ −1 (U), f ◦ φ〉 ∈ K(X). We obtain a K–homomorphism, which is necessarily injective, K(Y ) → K(X), also denoted φ ∗ .Note that in this case, we have: dim X ≥ dimY . As above, it is possible to checkthat, if X ≃ Y , then K(X) ≃ K(Y ), hence dimX = dim Y . Moreover, if P ∈ Xand Q = φ(P), then φ ∗ induces a map O Q,Y → O P,X , such that φ ∗ M Q,Y ⊂ M P,X .Also in this case, if φ is an isomorphism, then O Q,Y ≃ O P,X .Let now X ⊂ P n be a quasi–projective variety and φ : X → P m be a map.9.6. Proposition. φ is a morphism if and only if, for all P ∈ X, there existan open neighbourhood U P of P and n + 1 homogeneous polynomials F 0 , . . ., F mof the same degree, in K[x 0 , x 1 , . . ., x n ], such that, if Q ∈ U P , then φ(Q) =[F 0 (Q), . . ., F m (Q)]. In particular, for all Q ∈ U P , there exists an index i suchthat F i (Q) ≠ 0.Proof. “ ⇒” Let P ∈ X, Q = φ(P) and assume that Q ∈ U 0 . Then U := φ −1 (U 0 )is an open neighbourhood of P and we can consider the restriction φ| U : U → U 0 ,which is regular. Possibly after restricting U, using non–homogeneous coordinateson Y 0 , we can assume that φ| U = (F 1 /G 1 , . . ., F m /G m ), where (F 1 , G 1 ),. . ., (F m , G m ) are pairs of homogeneous polynomials of the same degree suchthat V P (G i ) ∩ U = ∅ for all index i. We can reduce the fractions F i /G i to acommon denominator F 0 , so that deg F 0 = deg F 1 = . . . = deg F m and φ| U =(F 1 /F 0 , . . ., F m /F 0 ) = [F 0 , F 1 , . . ., F m ], with F 0 (Q) ≠ 0 for Q ∈ U.“ ⇐” Possibly after restricting U P , we can assume F i (Q) ≠ 0 for all Q ∈ U Pand suitable i. Let i = 0: then φ| UP : U P → U 0 operates as follows: φ| UP (Q) =(F 1 (Q)/F 0 (Q), . . ., F m (Q)/F 0 (Q)), so it is a morphism by Proposition 9.3. Fromthis remark, one deduces that also φ is a morphism.□9.7. Examples.1. Let X ⊂ P 2 , X = V P (x 2 1 + x2 2 − x2 0 ), the projective closure of the unitary

circle. We define φ : X → P 1 byIntroduction to algebraic geometry 41[x 0 , x 1 , x 2 ] →{[x0 − x 2 , x 1 ] if (x 0 − x 2 , x 1 ) ≠ (0, 0);[x 1 , x 0 + x 2 ] if (x 1 , x 0 + x 2 ) ≠ (0, 0).φ is well–defined because on X x 2 1 = (x 0 − x 2 )(x 0 + x 2 ). Moreover(x 1 , x 0 − x 2 ) ≠ (0, 0) ⇔ [x 0 , x 1 , x 2 ] ∈ X \ {[1, 0, 1]},(x 0 + x 2 , x 1 ) ≠ (0, 0) ⇔ [x 0 , x 1 , x 2 ] ∈ X \ {[1, 0, −1]}.The map φ is the natural extension of the rational function f : X \{[1, 0, 1]} →K such that [x 0 , x 1 , x 2 ] → x 1 /(x 0 −x 2 ) (Example 8.9, 2). Now the point P[1, 0, 1],the centre of the stereographic projection, goes to the point at infinity of the lineV P (x 2 ).By geometric reasons φ is invertible and φ −1 : P 1 → X takes [λ, µ] to [λ 2 +µ 2 , 2λµ, λ 2 − µ 2 ] (note the connection with the Pitagorean triples!).Indeed: the line through P and [λ, µ, 0] has equation: µx 0 − λx 1 − µx 2 = 0.Its intersections with X are represented by the system:{ µx0 − λx 1 − µx 2 = 0x 2 1 + x2 2 − x2 0 = 0Assuming µ ≠ 0 this system is equivalent to the following:{ µx0 − λx 1 − µx 2 = 0µ 2 x 2 0 = µ2 (x 2 1 + x2 2 ) = (λx 1 + µx 2 ) 2 .Therefore, either x 1 = 0 and x 0 = x 2 , orthe required expression.{(µ 2 − λ 2 )x 1 − 2λµx 2 = 0µx 0 = λx 1 + µx 2, which gives2. Affine transformations.Let A = (a ij ) be a n × n–matrix with entries in K, let B = (b 1 , . . ., b n ) ∈ A nbe a point. The map τ A : A n → A n defined by (x 1 , . . ., x n ) → (y 1 , . . ., y n ), suchthat{y i = ∑ a ij x j + b i , i = 1, . . ., n,jis a regular map called an affine transformation of A n . In matrix notation τ Ais Y = AX + B. If A is of rank n, then τ A is said non–degenerate and is anisomorphism: the inverse map τ −1Ais represented by X = A−1 Y −A −1 B. More ingeneral, an affine transformation from A n to A m is a map represented in matrixform by Y = AX + B, where A is a n × m matrix and B ∈ A m . It is injective ifand only if rkA = n and surjective if and only if rkA = m.

42 MezzettiThe isomorphisms of an algebraic set X in itself are called automorphismsof X: they form a group for the usual composition of maps, denoted Aut X. IfX = A n , the non–degenerate affine transformations form a subgroup of Aut A n .If n = 1 and the characteristic of K is 0, then Aut A 1 coincides with this subgroup.In fact, let φ : A 1 → A 1 be an automorphism: it is represented by a polynomialF(x) such that there exists G(x) satisfying the condition G(F(t)) = t for allt ∈ A 1 , i.e. G(F(x)) = x in the polynomial ring K[x]. Then G ′ (F(x))F ′ (x) = 1,which implies F ′ (t) ≠ 0 for all t ∈ K, so F ′ (x) is a non–zero constant. Hence, Fis linear and G is linear too.If n ≥ 2, then Aut A n is not completely described. There exist non–linearautomorphisms of degree d, for all d. For example, for n = 2: let φ : A 2 → A 2be given by (x, y) → (x, y + P(x), where P is any polynomial of K[x]. Thenφ −1 : (x ′ , y ′ ) → (x ′ , y ′ − P(x ′ )). A very important open problem is the Jacobianconjecture, stating that a regular map φ : A n → A n is an automorphism if andonly if the Jacobian determinant | J(φ) | is a non-zero constant.3. Projective transformations.Let A be a (n+1)×(n+1)–matrix with entries in K. Let P[x 0 , . . ., x n ] ∈ P n :then [a 00 x 0 +. . .+a 0n x n , . . ., a n0 x 0 +. . .+a nn x n ] is a point of P n if and only if itis different from [0, . . ., 0]. So A defines a regular map τ : P n → P n if and only ifrkA = n+1. If rkA = r < n+1, then A defines a regular map whose domain is thequasi–projective variety P n \ P(kerA). If rkA = n + 1, then τ is an isomorphism,called a projective transformation. Note that the matrices λA, λ ∈ K ∗ , all definethe same projective transformation. So PGL(n + 1, K) := GL(n + 1, K)/K ∗ actson P n as the group of projective transformations.If X, Y ⊂ P n , they are called projectively equivalent if there exists a projectivetransformation τ : P n → P n such that τ(X) = Y .9.8. Theorem. Fundamental theorem on projective transformations.Let two (n+2)–tuples of points of P n in general position be fixed: P 0 , . . ., P n+1and Q 0 , . . ., Q n+1 . Then there exists one isomorphic projective transformation τof P n in itself, such that τ(P i ) = Q i for all index i.Proof. Put P i = [v i ], Q i = [w i ], i = 0, . . ., n+1. So {v 0 , . . ., v n } and {w 0 , . . ., w n }are two bases of K n+1 , hence there exist scalars λ 0 , . . ., λ n , µ 0 , . . ., µ n such thatv n+1 = λ 0 v 0 + . . . + λ n v n , w n+1 = µ 0 w 0 + . . . + µ n w n ,where the coefficients are all different from 0, because of the general positionassumption. We replace v i with λ i v i and w i with µ i w i and get two new bases, sothere exists a unique automorphism of K n+1 transforming the first basis in thesecond one and, by consequence, also v n+1 in w n+1 . This automorphism inducesthe required projective transformation on P n .□An immediate consequence of the above theorem is that projective subspacesof the same dimension are projectively equivalent. Also two subsets of P n formed

Introduction to algebraic geometry 43both by k points in general position are projectively equivalent if k ≤ n + 2. Ifk > n + 2, this is no longer true, already in the case of four points on a projectiveline. The problem of describing the classes of projective equivalence of k–tuplesof points of P n , for k > n + 2, is one the first problems of the classical invarianttheory. The solution in the case k = 4, n = 1 is given by the notion of cross–ratio.4. Let X ⊂ A n be an affine variety, then X F = X \ V (F) is isomorphic toa closed subset of A n+1 , i.e. to Y = V (x n+1 F − 1, G 1 , . . ., G r ), where I(X) =〈G 1 , . . ., G r 〉. Indeed, the following regular maps are inverse each other:φ : X F → Y such that (x 1 , . . ., x n ) → (x 1 , . . ., x n , 1/F(x 1 , . . ., x n )),ψ : Y → X F such that (x 1 , . . ., x n , x n+1 ) → (x 1 , . . ., x n ).Hence, X F is a quasi–projective variety contained in A n , not closed in A n , butisomorphic to a closed subset of another affine space.From now on, the term affine variety will denote a quasi–projective varietyisomorphic to some affine closed set.If X is an affine variety and precisely X ≃ Y , with Y ⊂ A n closed, thenO(X) ≃ O(Y ) = K[t 1 , . . ., t n ] is a finitely generated K–algebra. In particular,if K is algebraically closed and α is an ideal strictly contained in O(X), thenV (α) ⊂ X is non–empty, by the relative form of the Nullstellensatz. From thisobservation, we can deduce that the quasi–projective variety of next example isnot affine.5. A 2 \ {(0, 0)} is not affine.Set X = A 2 \ {(0, 0)}: first of all we will prove that O(X) ≃ K[x, y] = O(A 2 ),i.e. any regular function on X can be extended to a regular function on the wholeplane.Indeed: let f ∈ O(X): if P ≠ Q are points of X, then there exist polynomialsF, G, F ′ , G ′ such that f = F/G on a neighbourhood U P of P and f = F ′ /G ′ ona neighbourhood U Q of Q. So F ′ G = FG ′ on U P ∩ U Q ≠ ∅, which is open alsoin A 2 , hence dense. Therefore F ′ G = FG ′ in K[x, y]. We can clearly assume thatF and G are coprime and similarly for F ′ and G ′ . So by the unique factorizationproperty, it follows that F ′ = F and G ′ = G. In particular f admits a uniquerepresentation as F/G on X and G(P) ≠ 0 for all P ∈ X. Hence G has no zeroeson A 2 , so G = c ∈ K ∗ and f ∈ O(X).Now, the ideal 〈x, y〉 has no zeroes in X and is proper: this proves that X isnot affine.On the other hand, the following property holds: 9.9. Proposition. LetX ⊂ P n be quasi–projective. Then X admits an open covering by affine varieties.Proof. Let X = X 0 ∪ . . . ∪ X n be the open covering of X where X i = U i ∩ X= {P ∈ X|P[a 0 , . . ., a n ], a i ≠ 0}. So, fixed P, there exists an index i such thatP ∈ X i . We can assume that P ∈ X 0 : X 0 is open in some affine variety Y of A n(identified with U 0 ); set X 0 = Y \ Y ′ , where Y , Y ′ are both closed. Since P ∉ Y ′ ,

44 Mezzettithere exists F such that F(P) ≠ 0 and V (F) ⊃ Y ′ . So P ∈ Y \ V (F) ⊂ Y \ Y ′and Y \ V (F) is an affine open neighbourhood of P in Y \ Y ′ = X 0 ⊂ X.□6. The Veronese maps.Let n, d be positive integers; put N(n, d) = ( ) (n+dd − 1. Note that n+d)d isequal to the number of (monic) monomials of degree d in the variables x 0 , . . ., x n ,that is equal to the number of n+1–tuples (i 0 , . . ., i n ) such that i 0 +...+ i n = d,i j ≥ 0. Then in P N(n,d) we can use coordinates {v i0 ...i n}, where i 0 , . . ., i n ≥ 0 andi 0 + . . . + i n = d. For example: if n = 2, d = 2, then N(2, 2) = ( )42 − 1 = 5. In P5we can use coordinates v 200 , v 110 , v 101 , v 020 , v 011 , v 002 .For all n, d we define the map v n,d : P n → P N(n,d) such that [x 0 , . . ., x n ] →[v d00...0 , v d−1,10...0 , . . ., v 0...00d ] where v i0 ...i n= x i 00 x i 11 . . .x i nn : v n,d is clearly a morphism,its image is denoted V n,d and called the Veronese variety of type (n, d). Itis in fact the projective variety of equations:(∗){v i0 ...i nv j0 ...j n− v h0 ...h nv k0 ...k n, ∀i 0 + j 0 = h 0 + k 0 , i 1 + j 1 = h 1 + k 1 , . . .We prove this statement in the particular case n = d = 2; the general case issimilar.First of all, it is clear that the points of v n,d (P n ) satisfy the system (∗).Conversely, assume that P[v 200 , v 110 , . . .] ∈ P 5 satisfies the equations (∗), whichbecome:⎧v 200 v 020 = v1102v ⎪⎨ 200 v 002 = v1012v 002 v 020 = v0112v 200 v 011 = v 110 v 101⎪⎩ v 020 v 101 = v 110 v 011v 110 v 002 = v 011 v 101Then, at least one of the coordinates v 200 , v 020 , v 002 is different from 0.Therefore, if v 200 ≠ 0, then P = v 2,2 ([v 200 , v 110 , v 101 ]); if v 020 ≠ 0, thenP = v 2,2 ([v 110 , v 020 , v 011 ]); if v 002 ≠ 0, then P = v 2,2 ([v 101 , v 011 , v 002 ]). Note that,if two of these three coordinates are different from 0, then the points of P 2 foundin this way have proportional coordinates, so they coincide.We have also proved in this way that v 2,2 is an isomorphism between P 2 andV 2,2 , called the Veronese surface of P 5 . The same happens in the general case.If n = 1, v 1,d : P 1 → P d takes [x 0 , x 1 ] to [x d 0 , xd−1 0 x 1 , . . ., x d 1]: the image iscalled the rational normal curve of degree d, it is isomorphic to P 1 . If d = 3, wefind the skew cubic.Let now X ⊂ P n be a hypersurface of degree d: X = V P (F), withF =∑i 0 +...+i n =da i0 ...i nx i 00 . . .x i nn .

Then v n,d (X) ≃ X: it is the set of points{v i0 ...i n∈ P N(n,d) |Introduction to algebraic geometry 45∑i 0 +...+i n =da i0 ...i nv i0 ...i n= 0 and [v i0 ...i n] ∈ V n,d }.It coincides with V n,d ∩ H, where H is a hyperplane of P N(n,d) : a hyperplanesection of the Veronese variety.The Veronese surface V of P 5 enjoys a lot of interesting properties. Most ofthem follow from its property of being covered by a 2-dimensional family of conics,which are precisely the images via v 2,2 of the lines of the plane.To see this, we’ll use as coordinates in P 5 w 00 , w 01 , w 02 , w 11 , w 12 , w 22 , so thatv 2,2 sends [x 0 , x 1 , x 2 ] to the point of coordinates w ij = x i x j . With this choice ofcoordinates, the equations of V are obtained by annihilating the 2 × 2 minors ofthe symmetric matrix:⎛M = ⎝ w ⎞00 w 01 w 02w 01 w 11 w 12⎠w 02 w 12 w 22Let l be a line of P 2 of equation b 0 x 0 + b 1 x 1 + b 2 x 2 = 0. Its image is the set ofpoints of P 5 with coordinates w ij = x i x j , such that there exists a non-zero triple[x 0 , x 1 , x 2 ] with b 0 x 0 +b 1 x 1 +b 2 x 2 = 0. But this last equation is equivalent to thesystem:⎧⎨b 0 x 2 0 + b 1x 0 x 1 + b 2 x 0 x 2 = 0b⎩ 0 x 0 x 1 + b 1 x 2 1 + b 2x 1 x 2 = 0b 0 x 0 x 2 + b 1 x 1 x 2 + b 2 x 2 2 = 0It represents the intersection of V with the plane(∗){ b0 w 00 + b 1 w 01 + b 2 w 02 = 0b 0 w 01 + b 1 w 11 + b 2 w 12 = 0 ,b 0 w 02 + b 1 w 12 + b 2 w 22 = 0so v 2,2 (l) is a plane curve. Its degree is the number of points in its intersectionwith a general hyperplane in P 5 : this corresponds to the intersection in P 2 of lwith a conic (a hypersurface of degree 2). Therefore v 2,2 (l) is a conic.So the isomorphism v 2,2 transforms the geometry of the lines in the planein the geometry of the conics on the Veronese surface. In particular, given twodistinct points on V , there is exactly one conic contained in V and passing throughthem.From this observation it is easy to deduce that the secant lines of V , i.e. thelines meeting V at two points, are precisely the lines of the planes generated bythe conics contained in V , so that the (closure of the) union of these secant lines

46 Mezzetticoincides with the union of the planes of the conics of V . This union results to bethe cubic hypersurface defined by the equation⎛det M = det ⎝ w ⎞00 w 01 w 02w 01 w 11 w 12⎠ = 0.w 02 w 12 w 22Indeed a point of P 5 , of coordinates [w ij ] belongs to the plane of a conic containedin V if and only if there exists a non-zero triple [b 0 , b 1 , b 2 ] which is solution of thehomogeneous system (*).Let X, Y be quasi–projective varieties.9.10. Definition. The rational maps from X to Y are the germs of regular mapsfrom open subsets of X to Y , i.e. equivalence classes of pairs (U, φ), where U ≠ ∅is open in X and φ : U → Y is regular, with respect to the relation: (U, φ) ∼ (V, ψ)if and only if φ| U∩V = ψ| U∩V . The following Lemma guarantees that the abovedefined relation satisfies the transitive property.9.11. Lemma. Let φ, ψ : X → Y be regular maps between quasi-projective varieties.If φ| U = ψ| U for U ⊂ X open and non–empty, then φ = ψ.Proof. Let P ∈ X and consider φ(P), ψ(P) ∈ Y . There exists a hyperplane Hsuch that φ(P) ∉ H and ψ(P) ∉ H. Up to a projective transformation, we canassume that H = V P (x 0 ), so φ(P), ψ(P) ∈ U 0 . Set V = φ −1 (U 0 ) ∩ ψ −1 (U 0 ): anopen neighbourhood of P. Consider the restrictions of φ and ψ from V to Y ∩U 0 :they are regular maps which coincide on V ∩ U, hence their coordinates φ i , ψ i ,i = 1, . . ., n, coincide on V ∩ U. So φ i | V = ψ i | V . In particular φ(P) = ψ(P). □A rational map from X to Y will be denoted φ : X Y . As for rationalfunctions, the domain of definition of φ, dom φ, is the maximum open subset ofX such that φ is regular at the points of dom φ.The following proposition follows from the characterization of rational functionson affine varieties.9.12. Proposition. Let X, Y be affine algebraic sets, with Y closed in A n . Thenφ : X Y is a rational map if and only if φ = (φ 1 , . . ., φ n ), where φ 1 , . . ., φ n ∈K(X).□If X ⊂ P n , Y ⊂ P m , then a rational map X Y is assigned by giving m+1homogeneous polynomials of K[x 0 , x 1 , . . ., x n ] of the same degree, F 0 , . . ., F m , suchthat at least one of them is not identically zero on X.A rational map φ : X Y is called dominant if the image of X via φ isdense in X, i.e. if φ(U) = X, where U = dom φ. If φ : X Y is dominant

Introduction to algebraic geometry 47and ψ : Y Z is any rational map, then dom ψ ∩ Imφ ≠ ∅, so we can defineψ ◦ φ : X Z: it is the germ of the map ψ ◦ φ, regular on φ −1 (dom ψ ∩ Imφ).9.13. Definition. A birational map from X to Y is a rational map φ : X Ysuch that φ is dominant and there exists ψ : Y X, a dominant rational map,such that ψ ◦ φ = 1 X and φ ◦ ψ = 1 Y as rational maps. In this case, X and Y arecalled birationally equivalent or simply birational.If φ : X Y is a dominant rational map, then we can define the comorphismφ ∗ : K(Y ) → K(X) in the usual way: it is an injective K–homomorphism.9.14. Proposition. Let X, Y be quasi–projective varieties, u : K(Y ) → K(X)be a K–homomorphism. Then there exists a rational map φ : X Y such thatφ ∗ = u.Proof. Y is covered by open affine varieties Y α , α ∈ I (by Proposition 9.9): for allindex α, K(Y ) ≃ K(Y α ) (Prop. 8.8) and K(Y α ) ≃ K(t 1 , . . ., t n ), where t 1 , . . ., t ncan be interpreted as coordinate functions on Y α . Then u(t 1 ), . . ., u(t n ) ∈ K(X)and there exists U ⊂ X, non–empty open subset such that u(t 1 ), . . ., u(t n ) are allregular on U. So u(K[t 1 , . . ., t n ]) ⊂ O(U) and we can consider the regular mapu ♯ : U → Y α ֒→ Y . The germ of u ♯ gives a rational map X Y . It is possibleto check that this rational map does not depend on the choice of Y α and U. □9.15. Theorem. Let X, Y be quasi–projective varieties. The following are equivalent:(i) X is birational to Y ;(ii) K(X) ≃ K(Y );(iii) there exist non–empty open subsets U ⊂ X and V ⊂ Y such that U ≃ V .Proof.(i) ⇔ (ii) via the construction of the comorphism φ ∗ associated to φ and ofu ♯ , associated to u : K(Y ) → K(X). One checks that both constructions arefunctorial.(i) ⇒ (iii) Let φ : X Y , ψ : Y X be inverse each other. PutU ′ = dom φ and V ′ = dom ψ. By assumption, ψ ◦ φ is defined on φ −1 (V ′ )and coincides with 1 X there. Similarly, ψ ◦ φ is defined on ψ −1 (U ′ ) and equalto 1 Y . Then φ and ψ establish an isomorphism between the corresponding setsU := φ −1 (ψ −1 (U ′ )) and V := ψ −1 (φ −1 (V ′ )).(iii) ⇒ (ii) U ≃ V implies K(U) ≃ K(V ); but K(U) ≃ K(X) and K(V ) ≃K(Y ) (Prop.8.8), so K(X) ≃ K(Y ) by transitivity.□9.16. Corollary. If X is birational to Y , then dimX = dim Y . □9.17. Examples.

48 Mezzettia) The cuspidal cubic Y = V (x 3 − y 2 ) ⊂ A 2 .We have seen that Y is not isomorphic to A 1 , but in fact Y and A 1 arebirational. Indeed, the regular map φ : A 1 → Y , t → (t 2 , t 3 ), admits a rationalinverse ψ : Y A 1 , (x, y) → y x. ψ is regular on Y \ {(0, 0)}, ψ is dominant andψ◦φ = 1 A 1, φ◦ψ = 1 Y as rational maps. In particular, φ ∗ : K(Y ) → K(X) is a fieldisomorphism. Recall that K[Y ] = K[t 1 , t 2 ], with t 2 1 = t3 2 , so K(Y ) = K(t 1, t 2 ) =K(t 2 /t 1 ), because t 1 = (t 2 /t 1 ) 2 = t 2 2 /t2 1 = t3 1 /t2 1 and t 2 = (t 2 /t 1 ) 3 = t 3 2 /t3 1 = t3 2 /t2 2 ,so K(Y ) is generated by a unique transcendental element. Notice that φ and ψestablish isomorphisms between A 1 \ {0} and Y \ {(0, 0)}.b)Rational maps from P 1 to P n .Let φ : P 1 P n be rational: on some open U ⊂ P 1 ,φ([x 0 , x 1 ]) = [F 0 (x 0 , x 1 ), . . ., F n (x 0 , x 1 )],with F 0 , . . ., F n homogeneous of the same degree, without non–trivial commonfactors. Assume that F i (P) = 0 for a certain index i, with P = [a 0 , a 1 ]. ThenF i ∈ I h (P) = 〈a 1 x 0 −a 0 x 1 〉, i.e. a 1 x 0 −a 0 x 1 is a factor of F i . This remark impliesthat ∀ Q ∈ P 1 there exists i ∈ {0, . . ., n} such that F i (Q) ≠ 0, because otherwiseF 0 , . . ., F n would have a common factor of degree 1. Hence we conclude that φ isregular.c) Projections.Let φ : P n P m be given in matrix form by Y = AX, where A is a(m + 1) × (n + 1)-matrix, with entries in K. Then φ is a rational map, regular onP n \ P(KerA). Put Λ := P(KerA). If A = (a ij ), this means that Λ has cartesianequations⎧a 00 x 0 + . . . + a 0n x n = 0⎪⎨a 10 x 0 + . . . + a 1n x n = 0⎪⎩ . . .a m0 x 0 + . . . + a mn x n = 0The map φ has a geometric interpretation: it can be seen as the projectionof centre Λ to a complementar linear space. First of all, we can assume that rkA = m + 1, otherwise we replace P m with P(Im A); hence dimΛ = n − (m + 1).Consider first the case Λ : x 0 = . . . = x m = 0; we identify P m with thesubspace of P n of equations x m+1 = . . . = x n = 0, so Λ and P m are complementarsubspaces, i.e. Λ ∩ P m = ∅ and the linear span of Λ and P m is P n . Then, forQ ∈ P n \ Λ, φ(Q) = [x 0 , . . ., x m , 0, . . ., 0]: it is the intersection of P m with thelinear span of Λ and Q. In fact, if Q[a 0 , . . ., a n ] then ΛQ has equations{a i x j − a j x i = 0, i, j = 0, . . ., m (check!)so ΛQ ∩ P m has coordinates [a 0 , . . ., a m , 0, . . ., 0].

Introduction to algebraic geometry 49In the general case, if Λ = V P (L 0 , . . ., L m ), with L 0 , . . ., L m linearly independentforms, we can identify P m with V P (L m+1 , . . ., L n ), where L 0 , . . ., L m ,L m+1 , . . ., L n is a basis of (K n+1 ) ∗ . Then L 0 , . . ., L m can be interpreted as coordinatefunctions on P m .If m = n − 1, then Λ is a point P and φ, often denoted π P , is the projectionfrom P to a hyperplane not containing P.d)Rational and unirational varieties.A quasi–projective variety X is called rational if it is birational to a projectivespace P n (or equivalently to A n ). By Theorem 9.15, X is rational if and only ifK(X) ≃ K(P n ) = K(x 1 , . . ., x n ) for some n, i.e. if K(X) is an extension of Kgenerated by a transcendence basis. In an equivalent way, X is rational if thereexists a rational map φ : P n X which is dominant and is an isomorphismif restricted to a suitable open subset U ⊂ P n . Hence X admits a birationalparametrization by n parameters.A weaker notion is that of unirational variety: X is unirational if there existsa dominant rational map P n X i.e. if K(X) is contained in the quotient field ofa polynomial ring. Hence X can be parametrized, but not necessarily genericallyone–to–one.It is clear that, if X is rational, then it is unirational. The converse implicationhas been an important open problem, up to 1971, when it has been solved inthe negative, for varieties of dimension ≥ 3 (Clemens–Griffiths and Iskovskih–Manin). The result is true for curves (Theorem of Lüroth, 1880) and for surfacesif charK = 0 (Theorem of Castelnuovo, 1894).As an example of rational variety, let us consider the following quadric ofmaximal rank in P 3 : X = V P (x 0 x 3 − x 1 x 2 ): an irreducible hypersurface of degree2. Let π P : P 3 P 2 be the projection of centre P[1, 0, 0, 0], such thatπ P ([y 0 , y 1 , y 2 , y 3 ]) = [y 1 , y 2 , y 3 ]. The restriction of π P to X is a rational mapπ˜P : X P 2 , regular on X \ {P }. π˜P has a rational inverse: indeed considerthe rational map ψ : P 2 X, [y 1 , y 2 , y 3 ] → [y 1 y 2 , y 1 y 3 , y 2 y 3 , y3]. 2 The equationof X is satisfied by the points of ψ(P 2 ): (y 1 y 2 )y3 2 = (y 1y 3 )(y 2 y 3 ). ψ is regular onP 2 \ V P (y 1 y 2 , y 3 ). Let us compose ψ and ˜π P :[y 0 , . . ., y 3 ] ∈ X π P→ [y 1 , y 2 , y 3 ] → ψ [y 1 y 2 , y 1 y 3 , y 2 y 3 , y3];2y 1 y 2 = y 0 y 3 implies ψ ◦ π P = 1 X . In the opposite order:[y 1 , y 2 , y 3 ] → ψ [y 1 y 2 , y 1 y 3 , y 2 y 3 , y3 2 ] π P→ [y 1 y 3 , y 2 y 3 , y3 2 ] = [y 1, y 2 , y 3 ].So X is birational to P 2 hence it is a rational surface.e) A birational non–regular map from P 2 to P 2 .

50 MezzettiThe following rational map is called the standard quadratic map:Q : P 2 P 2 , [x 0 , x 1 , x 2 ] → [x 1 x 2 , x 0 x 2 , x 0 x 1 ].Q is regular on U := P 2 \ {A, B, C}, where A[1, 0, 0], B[0, 1, 0], C[0, 0, 1] are thefundamental points (see Fig. 2)Let a be the line through B and C: a = V P (x 0 ), and similarly b = V P (x 1 ),c = V P (x 2 ). Then Q(a) = A, Q(b) = B, Q(c) = C. Outside these three lines Q isan isomorphism. Precisely, put U ′ = P 2 \ {a ∪ b ∪ c}; then Q : U ′ → P 2 is regular,the image is U ′ and Q −1 : U ′ → U ′ coincides with Q. Indeed,[x 0 , x 1 , x 2 ] Q → [x 1 x 2 , x 0 x 2 , x 0 x 1 ] Q → [x 2 0x 1 x 2 , x 0 , x 2 1x 2 , x 0 x 1 x 2 2].So Q ◦ Q = 1 P 2 as rational map, hence Q is birational and Q = Q −1 .– Fig.2 –The set of the birational maps P 2 P 2 is a group, called the Cremonagroup. At the end of XIX century, Max Noether proved that the Cremona groupis generated by PGL(3, K) and by the single standard quadratic map above. Theanalogous groups for P n , n ≥ 3, are much more complicated and a completedescription is still unknown.Exercises to §9.1. Let φ : A 1 → A n be the map defined by t → (t, t 2 , . . ., t n ).a) Prove that φ is regular and describe φ(A 1 );b) prove that φ : A 1 → φ(A 1 ) is an isomorphism;c) give a description of φ ∗ and φ −1∗ .2. Let f : A 2 → A 2 be defined by: (x, y) → (x, xy).

Introduction to algebraic geometry 51a) Describe f(A 2 ) and prove that it is not locally closed in A 2 .b) Prove that f(A 2 ) is a constructible set in the Zariski topology of A 2 (i.e.a finite union of locally closed sets).3. Prove that the Veronese variety V n,d is not contained in any hyperplane ofP N(n,d) .4. Let GL n (K) be the set of invertible n × n matrices with entries in K.Prove that GL n (K) can be given the structure of an affine variety.5. Show the unicity of the projective transformation τ of Theorem 9.8.6. Let φ : X → Y be a regular map and φ ∗ its comorphism. Prove that thekernel of φ ∗ is the ideal of φ(X) in O(Y ). Deduce that φ is dominant if and onlyif φ ∗ is injective.7. Prove that O(X F ) is isomorphic to O(X) f , where X is an affine algebraicset, F a polynomial and f the function on X defined by F.10. Products of quasi–projective varieties.Let P n , P m be projective spaces over the same field K. The cartesian productP n × P m is simply a set: we want to define an injective map from P n × P m to asuitable projective space, so that the image be a projective variety, which will beidentified with our product.Let N = (n(+1)(m + 1) − 1 and define σ : P n × P m → P N in the followingway: ([x 0 , . . ., x n ], [y 0 , . . ., y m ]) → [x 0 y 0 , x 0 y 1 , . . ., x i y j , . . ., x n y m ]. Using coordinatesw ij , i = 0, . . ., n, j = 0, . . ., m, in P N , σ is given by{w ij = x i y j , i = 0, . . ., n, j = 0, . . ., m.It is easy to observe that σ is a well–defined map.If P n = P(V ) and P m = P(W), note that P N ≃ P(V ⊗ W) and σ is inducedby the natural map V × W → V ⊗ W.Let Σ n,m (or simply Σ) denote the image σ(P n × P m ).10.1. Proposition. σ is injective and Σ n,m is a closed subset of P N .Proof. If σ([x], [y]) = σ([x ′ ], [y ′ ]), then there exists λ ≠ 0 such that x ′ i y′ j = λx iy jfor all i, j. In particular, if x h ≠ 0, y k ≠ 0, then also x ′ h ≠ 0, y′ k ≠ 0, and for all ix ′ i = λ y kyx ′ i , so [x 0 , . . ., x n ] = [x ′ 0, . . ., x ′ n]. Similarly for the second point.kTo prove the second assertion, I claim: Σ n,m is the closed set of equations:(∗){w ij w hk = w ik w hj , i, h = 0, . . ., n; j, k = 0 . . ., m.It is clear that if [w ij ] ∈ Σ, then it satisfies (*). Conversely, assume that [w ij ]

52 Mezzettisatisfies (*) and that w αβ ≠ 0. Then[w 00 , . . ., w ij , . . ., w nm ] = [w 00 w αβ , . . ., w ij w αβ , . . ., w nm w αβ ] == [w 0β w α0 , . . ., w iβ w αj , . . ., w nβ w αm ] == σ([w 0β , . . ., w nβ ], [w α0 , . . ., w αm ]).σ is called the Segre map and Σ n,m the Segre variety or biprojective space. Notethat Σ is covered by the affine open subsets Σ ij = Σ ∩ W ij , where W ij = P N \V P (w ij ). Moreover Σ ij = σ(U i × V j ), with U i × V j ≃ A n+m .10.2. Proposition. σ| Ui ×V j: U i × V j ≃ A n+m → Σ ij is an isomorphism ofvarieties.Proof. Assume by simplicity i = j = 0. Choose non–homogeneous coordinates onU 0 : u i = x i /x 0 and on V 0 : v j = y j /y 0 . So u 1 , . . .u n , v 1 , . . ., v m are coordinates onU 0 × V 0 . Take non–homogeneous coordinates also on W 00 : z ij = w ij /w 00 . Usingthese coordinates we have:σ| Ui ×V j:(u 1 , . . .u n , v 1 , . . ., v m ) → (v 1 , . . ., v m , u 1 , u 1 v 1 , . . ., u 1 v m , . . ., u n v m )||([1, u 1 , . . ., u n ], [1, v 1 , . . ., v m ])i.e. σ(u 1 , . . ., v m ) = (z 01 , . . ., z nm ), where⎧⎨z i0 = u i , if i = 1, . . ., n;z 0j = v j , if j = 1, . . ., m;⎩z ij = u i v j = z i0 z 0j otherwise.Hence σ| U0 ×V 0is regular.The inverse map takes (z 01 , . . ., z nm ) to (z 10 , . . ., z n0 , z 01 , . . ., z 0m ), so it is alsoregular.□□10.3. Corollary. P n × P m is irreducible and birational to P n+m .Proof. The first assertion follows from Ex.5, Ch.6, considering the covering of Σby the open subsets Σ ij . For the second assertion, by Theorem 9.15, it is enoughto note that Σ n,m and P n+m contain isomorphic open subsets, i.e. Σ ij and A n+m .□From now on, we shall identify P n × P m with Σ n,m . If X ⊂ P n , Y ⊂ P m areany quasi–projective varieties, then X × Y will be automatically identified withσ(X × Y ) ⊂ Σ.

Introduction to algebraic geometry 5310.4. Proposition. If X and Y are projective varieties (resp. quasi–projectivevarieties), then X × Y is projective (resp. quasi–projective).Proof.σ(X × Y ) = ⋃ i,j(σ(X × Y ) ∩ Σ ij ) == ⋃ i,j(σ(X × Y ) ∩ (U i × V j )) == ⋃ i,j(σ((X ∩ U i ) × (Y ∩ V j ))).If X and Y are projective varieties, then X ∩U i is closed in U i and Y ∩V j is closedin V j , so their product is closed in U i × V j ; since σ| Ui ×V jis an isomorphism, alsoσ(X × Y ) ∩ Σ ij is closed in Σ ij , so σ(X × Y ) is closed in Σ, by Lemma 8.3.If X, Y are quasi–projective, the proof is similar. As for the irreducibility, seeExercise 10.1.□10.5. Example. P 1 × P 1σ : P 1 × P 1 → P 3 is given by {w ij = x i y j , i = 0, 1, j = 0, 1. Σ has only onenon–trivial equation: w 00 w 11 − w 01 w 10 , hence Σ is a quadric. The equation of Σcan be written as(∗)∣ w ∣00 w 01 ∣∣∣= 0.w 10 w 11Σ contains two families of special closed subsets parametrized by P 1 , i.e.{σ(P × P 1 )} P ∈P1 and {σ(P 1 × Q)} Q∈P 1.If P[a 0 , a 1 ], then σ(P × P 1 ) is given by the equations:⎧w 00 = a 0 y 0⎪⎨w 01 = a 0 y 1⎪⎩w 10 = a 1 y 0w 11 = a 1 y 1hence it is a line. Cartesian equations of σ(P × P 1 ) are:{a1 w 00 − a 0 w 10 = 0a 1 w 01 − a 0 w 11 = 0;they express the proportionality of the rows of the matrix (*) with coefficients[a 1 , −a 0 ]. Similarly, σ(P 1 × Q) is the line of equations{ a1 w 00 − a 0 w 01 = 0a 1 w 10 − a 0 w 11 = 0.

54 MezzettiHence Σ contains two families of lines: two lines of the same family are clearlydisjoint while two lines of different families intersect at one point (σ(P, Q)). Conversely,through any point of Σ there pass two lines, one for each family. Notethat Σ is exactly the quadric surface of Example 9.17, d) and that the projectionof centre [1, 0, 0, 0] realizes an explicit birational map between P 1 × P 1 and P 2 .Exercises to §10.1. Using Ex. 5 of §6, prove that, if X ⊂ P n , Y ⊂ P m are irreducible projectivevarieties, then X × Y is irreducible.2. (*) Let X ⊂ A n , Y ⊂ A n . Show that X ∩ Y ≃ (X × Y ) ∩ ∆ A n, where ∆ Anis the diagonal subvariety.3. Let L, M, N be the following lines in P 3 :L : x 0 = x 1 = 0, M : x 2 = x 3 = 0, N : x 0 − x 2 = x 1 − x 3 = 0.Let X be the union of lines meeting L, M and N: write equations for X anddescribe it: is it a projective variety? If yes, of what dimension and degree?4. Let X, Y be quasi–projective varieties, identify X × Y with its image viathe Segre map. Check that the two projection maps X ×Y p 1→ X, X ×Y p 2→ Y areregular. (Hint: use the open covering of the Segre variety by the Σ ij ’s.)11. The dimension of an intersection.Our aim in this section is to prove the following theorem:11.1. Theorem. Let K be an algebraically closed field. Let X, Y ⊂ P n bequasi–projective varieties. Assume that X ∩ Y ≠ ∅. Then if Z is any irreduciblecomponent of X ∩ Y , then dimZ ≥ dim X + dim Y − n.The proof uses in an essential way the Krull’s principal ideal theorem (see forinstance Atiyah–MacDonald [1]).The proof of Theorem 11.1 will be divided in three steps. Note first that wecan assume that X and Y both intersect U 0 ≃ A n , so, possibly after restrictingX and Y , we may work with closed subsets of the affine space. Put r = dim X,s = dim Y .Step 1. Assume that X = V (F) is an irreducible hypersurface, with F irreduciblepolynomial of K[x 1 , . . ., x n ]. The irreducible components of X ∩ Y correspond,by the Nullstellensatz, to the minimal prime ideals containing I(X ∩ Y )in K[x 1 , . . ., x n ]. Let me recall that I(X ∩ Y )= √ I(X) + I(Y )= √ 〈I(Y ), F 〉. Sothose prime ideals are the minimal ones over 〈I(Y ), F 〉. They correspond bijectivelyto minimal prime ideals containing 〈f〉 in O(Y ), where f is the regularfunction on Y defined by F. We distinguish two cases:- if Y ⊂ X = V (F), then f = 0 and Y ∩ X = Y ; s = dimY > r + s − n =(n − 1) + s − n. So the theorem is true.

Introduction to algebraic geometry 55- if Y ⊄ X, then f ≠ 0, moreover f is not invertible, otherwise X ∩ Y = ∅:hence the minimal prime ideals over 〈f〉 in O(Y ) have all height one, so for all Z,irreducible component of X ∩ Y , dim Z = dimY − 1 = r + s − n (Theorem 7.7).Step 2. Assume that I(X) is generated by n − r polynomials (where n − r isthe codimension of X): I(X) = 〈F 1 , . . ., F n−r 〉. Then we can argue by inductionon n − r: we first intersect Y with V (F 1 ), whose irreducible components are allhypersurfaces, and apply Step 1: all irreducible components of Y ∩ V (F 1 ) havedimension either s or s − 1. Then we intersect each of these components withV (F 2 ), and so on.Step 3. We use the isomorphism ψ : X ∩ Y ≃ (X × Y ) ∩ ∆ An (see Ex.2,§10). ψ preserves the irreducible components and their dimensions, so we considerinstead of X and Y , the algebraic sets X ×Y and ∆ A n, contained in A 2n . We havedim X ×Y = r+s (Proposition 7.10). ∆ An is a linear subspace of A 2n , so it satisfiesthe assumption of Step 2. Hence, for all Z we have: dim Z ≥ (r + s) + n − 2n =r + s − n.□The above theorem can be seen as a generalization of the Grassmann relationfor linear subspaces. It is not an existence theorem, because it says nothing aboutX ∩ Y being non–empty. But for projective varieties, the following more preciseversion of the theorem holds:11.2. Theorem. Let X, Y ⊂ P n be projective varieties of dimensions r, s. Ifr + s − n ≥ 0, then X ∩ Y ≠ ∅.Proof. Let C(X), C(Y ) be the affine cones associated to X and Y . Then C(X) ∩C(Y ) is certainly non–empty, because it contains the origin O(0, 0, . . ., 0). Assumewe know that C(X) has dimension r + 1 and C(Y ) has dimension s + 1: thenby Theorem 11.1 all irreducible components Z of C(X) ∩ C(Y ) have dimension≥ (r + 1) + (s + 1) − (n+1) = r + s − n + 1 ≥ 1, hence Z contains points differentfrom O. These points give raise to points of P n belonging to X ∩ Y . It remains toshow:11.3. Proposition. Let Y ⊂ P n be a projective variety.Then dim Y = dimC(Y ) − 1. If S(Y ) denotes the homogeneous coordinatering, hence also dimY = dim S(Y ) − 1.Proof. Let p : A n+1 \ {O} → P n be the canonical map. Let us recall that C(Y ) =p −1 (Y )∪{O}. Assume that Y 0 := Y ∩U 0 ≠ ∅ and consider also C(Y 0 ) = p −1 (Y 0 )∪{O}. Then we have:C(Y 0 ) = {(λ, λa 1 , . . ., λa n ) | λ ∈ K, (a 1 , . . ., a n ) ∈ Y 0 }.So we can define a birational map between C(Y 0 ) and Y 0 × A 1 as follows:(y 0 , y 1 , . . ., y n ) ∈ C(Y 0 ) → ((y 1 /y 0 , . . ., y n /y 0 ), y 0 ) ∈ Y 0 × A 1 ,

56 Mezzetti((a 1 , . . ., a n ), λ) ∈ Y 0 × A 1 → (λ, λa 1 , . . ., λa n ) ∈ C(Y 0 ).Therefore dim C(Y 0 ) = dim(Y 0 × A 1 ) = dimY 0 + 1. To conclude, it is enough toremark that dim Y = dimY 0 and dim C(Y ) = dim C(Y 0 ) = dim S(Y ). □11.4. Corollaries.1. If X, Y ⊂ P 2 are projective curves over an algebraically closed field, thenX ∩ Y ≠ ∅.2. P 1 × P 1 is not isomorphic to P 2 .Proof. 1. is a straightforward application of Theorem 11.2. To prove 2., assumeby contradiction that φ : P 1 × P 1 → P 2 is an isomorphism. If L, L ′ are skew lineson P 1 × P 1 , then φ(L), φ(L ′ ) are rational disjoint curves of P 2 , which contradicts1.If X, Y ⊂ P n are varieties of dimensions r, s, then r + s − n is called theexpected dimension of X ∩ Y . If all irreducible components Z of X ∩ Y have theexpected dimension, then we say that the intersection X ∩ Y is proper or that Xand Y intersect properly.For example, two plane projective curves X, Y intersect properly if they don’thave any common irreducible component. In this case, it is possible to previewthe number of points of intersections. Precisely, it is possible to associate to everypoint P ∈ X ∩ Y a number i(P), called the multiplicity of intersection of X andY at P, in such a way that ∑ P ∈X∩Y i(P) = dd′ , where d is the degree of X andd ′ is the degree of Y . This result is known as Theorem of Bézout, and is the firstresult of the branch of algebraic geometry called Intersection Theory. For a proofof the Theorem of Bézout, see for instance the classical book of Walker [8], or thebook of Fulton on Algebraic Curves [5].Let X be a closed subvariety of P n (resp. of A n ) of codimension r. X is calleda complete intersection if I h (X) (resp. I(X)) is generated by r polynomials.Hence, if X is a complete intersection of codimension r, then X is certainly theintersection of r hypersurfaces. Conversely, if X is intersection of r hypersurfaces,then, by Theorem 11.1, using induction, we deduce that dimX ≥ n − r; alsoassuming equality, we cannot conclude that X is a complete intersection, butsimply that I(X) is the radical of an ideal generated by r polynomials.11.5. Example. Let X ⊂ P 3 be the skew cubic. The homogeneous ideal of Xcontains the three polynomials F 1 , F 2 , F 3 , the 2 × 2–minors of the matrixM =( )x0 x 1 x 2,x 1 x 2 x 3which are linearly independent polynomials of degree 2. Note that I h (X) does notcontain any linear polynomial, because X is not contained in any hyperplane, andthat the homogeneous component of minimal degree 2 of I h (X) is a vector space

Introduction to algebraic geometry 57of dimension 3. Hence I h (X) cannot be generated by two polynomials, i.e. X isnot a complete intersection.Nevertheless, X is the intersection of the surfaces V P (F), V P (G), where∣ F =∣ x ∣ 0 x 1 ∣∣∣ x 0 x 1 x 2 ∣∣∣∣∣and G =x 1 x 2 x 1 x 2 x 3 .∣ x 2 x 3 x 0In fact: G = x 0 F − x 3 (x 0 x 3 − x 1 x 2 ) + x 2 (x 1 x 3 − x 2 2 ). If P[x 0, . . ., x 3 ] ∈ V P (F) ∩V P (G), then P is a zero of x 0 x 2 3 − 2x 1x 2 x 3 + x 3 2 , and also ofx 2 (x 0 x 2 3 − 2x 1x 2 x 3 + x 3 2 ) = x2 1 x2 3 − 2x 1x 2 2 x 3 + x 4 2 = (x 1x 3 − x 2 2 )2 .Hence P is a zero also of x 1 x 3 − x 2 2 . So P annihilates x 3(x 0 x 3 − x 1 x 2 ). If x 3 =0, then also x 2 = 0 and x 1 = 0: P = [1, 0, 0, 0] ∈ X. If x 3 ≠ 0, then P ∈V P (F 1 , F 2 , F 3 ) = X.The situation is that the skew cubic X is the set-theoretic intersection of aquadric and a cubic, which are tangent along X, so their intersection is X countedwith multiplicity 2.This example motivates the following definition: X is a set–theoretic completeintersection if codimX = r and the ideal of X is the radical of an ideal generated byr polynomials. It is an open problem if all irreducible curves of P 3 are set–theoreticcomplete intersections. For more details, see [4].Exercises to §11.1. Let X ⊂ P 2 be the union of three points not lying on a line. Prove thatthe homogeneous ideal of X cannot be generated by two polynomials.12. Complete varieties.We work over an algebraically closed field K.12.1. Definition. Let X be a quasi–projective variety. X is complete if, forany quasi–projective variety Y , the natural projection on the second factor p 2 :X × Y → Y is a closed map. (Note that both projections p 1 , p 2 are morphisms:see Exercise 4 to §10.)Example. The affine line A 1 is not complete: let X = Y = A 1 , p 2 :A 1 × A 1 = A 2 → A 1 is the map such that (x 1 , x 2 ) → x 2 . Then Z := V (x 1 x 2 − 1)is closed in A 2 but p 2 (Z) = A 1 \ {O} is not closed.12.2. Proposition. (i) If f : X → Y is a regular map and X is complete, thenf(X) is a closed complete subvariety of Y .(ii) If X is complete, then all closed subvarieties of X are complete.

58 MezzettiProof. (i) Let Γ f ⊂ X ×Y be the graph of f: Γ f = {(x, f(x)) | x ∈ X}. It is clearthat f(X) = p 2 (Γ f ), so to prove that f(X) is closed it is enough to check that Γ f isclosed in X ×Y . Let us consider the diagonal of Y : ∆ Y = {(y, y) | y ∈ Y } ⊂ Y ×Y .If Y ⊂ P n , then ∆ Y = ∆ P n∩(Y ×Y ), so it is closed because ∆ Pn is the closed subsetdefined in Σ n,n by the equations w ij − w ji = 0, i, j = 0, . . ., n. There is a naturalmap f ×1 Y : X ×Y → Y ×Y , (x, y) → (f(x), y), such that (f ×1 Y ) −1 (∆ Y ) = Γ f .It is easy to see that f × 1 Y is regular, so Γ f is closed.Let now Z be any variety and consider p 2 : f(X) × Z → Z and the regularmap f × 1 Z : X × Z → f(X) × Z. There is a commutative diagram:X × Zp ′ 2−→Z↓ f×1 Z ր p 2f(X) × ZIf T ⊂ f(X) × Z, then (f × 1 Z ) −1 (T) is closed and p 2 (T) = p ′ 2 ((f × 1 Z) −1 (T)) isclosed because X is complete. We conclude that f(X) is complete.(ii) Let T ⊂ X be a closed subvariety and Y be any variety. We have to provethat p 2 : T × Y → Y is closed. If Z ⊂ T × Y is closed, then Z is closed also inX × Y , hence p 2 (Z) is closed because X is complete.□12.3. Corollaries.1. If X is a complete variety, then O(X) ≃ K.2. If X is an affine complete variety, then X is a point.Proof. 1. If f ∈ O(X), f can be interpreted as a regular map f : X → A 1 . ByProposition 12.2, (i), f(X) is a closed complete subvariety of A 1 , which is notcomplete. Hence f(X) is a point, so f ∈ K.2. By 1., O(X) ≃ K. But O(X) ≃ K[x 1 , . . ., x n ]/I(X), hence I(X) ismaximal. By the Nullstellensatz, X is a point.□12.4. Theorem. Let X be a projective variety. Then X is complete.Proof. (sketch, see Šafarevič [7].)1. It is enough to prove that p 2 : P n × A m → A m is closed, for all n, m. Thiscan be observed by using the local character of closedness and the affine opencoverings of quasi–projective varieties.2. If x 0 , . . ., x n are homogeneous coordinates on P n and y 1 , . . ., y m are coordinateson A m , then any closed subvariety of P n × A m can be characterized as theset of common zeroes of a set of polynomials in the variables x 0 , . . ., x n , y 1 , . . ., y m ,homogeneous in the first group of variables x 0 , . . ., x n .3. Let Z ⊂ P n × A m be closed. Then Z is the set of solutions of a system ofequations{G i (x 0 , . . ., x n ; y 1 , . . ., y m ) = 0, i = 1, . . ., t

Introduction to algebraic geometry 59where G i is homogeneous in the x’s. A point P(y 1 , . . ., y m ) is in p 2 (Z) if and onlyif the system{G i (x 0 , . . ., x n ; y 0 , . . ., y m ) = 0, i = 1, . . ., thas a solution in P n , i.e. if the ideal of K[x 0 , . . ., x n ] generated by G 1 (x; y),. . .,G t (x; y) has at least one zero in P n . Hencep 2 (Z) = {(y 1 , . . ., y m )| ∀ d ≥ 1 〈G 1 (x; y), . . ., G t (x; y)〉 ⊅ K[x 0 , . . ., x n ] d } == ⋂ d≥1{(y 1 , . . ., y m )| 〈G 1 (x; y), . . ., G t (x; y)〉 ⊅ K[x 0 , . . ., x n ] d }.Let {M α } α∈I be the set of the monomials of degree d in K[x 0 , . . ., x n ]; let d i =deg G i (x; y); let {N β i } be the set of the monomials of degree d − d i; let finallyT d = {(y 1 , . . ., y m )| 〈G 1 (x; y), . . ., G t (x; y)〉 ⊅ K[x 0 , . . ., x n ] d }.Then P(y 1 , . . ., y m ) ∉ T d if and only if M α = ∑ i G i(x; y)F i,α (x 0 , . . ., x n ), forall α and for suitable polynomials F i,α homogeneous of degree d −d i . So P ∉ T d ifand only if for all α M α is a linear combination of the polynomials {G i (x; y)N β i },i.e. the matrix A of the coefficients of the polynomials G i (x; y)N β i with respect tothe basis {M α } has maximal rank ( )n+dd . So Td is the set of zeroes of the minorsof a fixed order of the matrix A, hence it is closed.□12.5. Corollary. Let X be a projective variety. Then O(X) ≃ K.12.6. Corollary. Let X be a projective variety, φ : X → Y be any regular map.Then φ(X) is a projective variety. In particular, if X ≃ Y , then Y is projective.References.[1] M. Atiyah - I. MacDonald, Introduction to Commutative Algebra, Addison- Wesley (1969)[2] J. Harris, Algebraic Geometry, Springer (1992)[3] R. Hartshorne, Algebraic Geometry, Springer (1977)[4] E. Kunz, Introduction to Commutative Algebra and Algebraic Geometry,Birkhäuser (1980)[5] W. Fulton, Algebraic Curves, Benjamin (1969)[6] S. Lang, Algebra, 2 nd ed., Addison-Wesley (1984)[7] I. Šafarevič, Basic Algebraic Geometry, Springer (1974)[8] R. Walker, Algebraic Curves, Springer (1978)[9] J. Semple - L. Roth, Algebraic Geometry, 1957[10] P. Samuel, O. Zariski, Commutative algebra (2 vol.), Van Nostrand, 1958[11] J. Dieudonné, Cours de géométrie algébrique, 2 / Précis de géométriealgébrique élémentaire, Presses Universitaires de France, 1974