Galois Theory and Noether's Problem Meredith Blue ... - MAA Sections

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Theorem 8 Let f(x) = min Q (α) for some α ∈ C. Suppose g(x) is a polynomialwith rational coefficients such that g(α) = 0. Then, g(x) = f(x)k(x)for some polynomial k(x).Proof: By the definition of min Q (α), deg(g(x)) ≥ deg(f(x)). Using theEuclidean algorithm for polynomials, we can writeg(x) = q(x)f(x) + r(x)where r(x) = 0 or deg(r(x)) < deg(f(x)). Plugging in α for x we haveg(α) = q(α)f(α) + r(α)g(α) = f(α) = 0, so r(α) = 0. Since deg(f(x)) is minimal with respect tohaving α as a root, r(x) = 0. Thus, g(x) = q(x)f(x).q.e.d.Galois is famous for his result that polynomials of degree 5 are not solvableby radicals, that is there can be no formula for solving all degree 5polynomials. In light of this we are interested in fields K ⊃ Q where Kis the smallest field containing all roots of a given polynomial with rationalcoefficients. These fields are called splitting fields:Definition 9 A field K ⊃ Q is a splitting field for the polynomial f(x) if1. K contains all roots of f(x).2. If L is another field containing all roots of f(x), then L ⊆ K.Galois studied these extension fields using automorphisms:Definition 10 A field automorphism is a mapping φ : F → F where Fis a field satisfying:φ(xy) = φ(x)φ(y)andφ(x + y) = φ(x) + φ(y)Let L/Q be a field extension. Let G be the set of automorphisms on Lthat leave Q fixed, that is:σ ∈ G if and only if σ : L → L is a field automorphism and σ(q) = q forall q ∈ Q.4
Theorem 11 If L/Q is a field extension, then the set, G(G, Q) of automorphismson L leaving Q fixed is a group with binary operation composition ofmappings.Let’s look at this another way. Suppose G is a group of automorphsimson a field F . We are interested in the subset F G ⊂ F that is fixed under allautomorphisms σ ∈ G. An easy exercise shows that this F G is in fact a field.Definition 12 Let G be the set of automorphisms on a field L. The set ofelements x ∈ L such that σ(x) = x for all x ∈ L and all σ ∈ G is called thefixed field of G. We denote this L G .Again, we look at some examples to observe that G(L, Q) may be differentfrom G(K, Q) if L and K are different extension fields of Q.Example: Consider Q( √ 2). Let G + G(Q( √ 2, Q) be the set of automorphismsof Q( √ 2) that leave Q fixed.There are 2 automorphisms of Q( √ 2) that leave Q fixed, the identity and:σ( √ 2) = − √ 2 σ(q) = q ∀ q ∈ QHence, the fixed field of G, Q( √ 2) G = Q.Example: Now consider Q( 3√ 2) ⊂ R. Let H be the set of automorphismsof Q( 3√ 2) that leave Q fixed.Notice that ( 3√ 2) is a root of x 3 − 2. Suppose τ is an automorphism ofQ( 3√ 2). Consider τ( 3√ 2 3 − 2).τ(0) = τ( 3√ 2 3 − 2) = (τ( 3√ 2)) 3 − 2Thus, any automorphism of Q( 3√ 2) leaving Q fixed will take 3√ 2 to anotherroot of x 3 − 2. Since there are no other real roots, all automorphisms in Hmust fix 3√ 2 also.Thus the fixed field of G is Q( 3√ 2).Definition 13 A field extension L/Q is Galois with group G if and onlyif G is the group of automorphisms of L leaving Q fixed and L G = Q.Let us now restate the Galois extension condition for the simple algebraicextension L = Q(α) for some α ∈ C. Let f(x) = min Q (α). In this case,L/Q is Galois if and only if L contains all roots of f(x). This is easily seenin the two examples above. Q( 3√ 2) only contains one root of f(x) = x 3 − 2,whereas Q( √ 2) contains both roots of f(x) = x 2 −2. This is easily describedin terms of splitting fields:5
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Page 3: in 2 variables over Q. These are in
Page 8 and 9: andσ(x 0 x 1 ) = σ(x 0 )σ(x 1 )

Galois Theory and Noether's Problem Meredith Blue ... - MAA Sections

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