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Section 14.4

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Section 14.4 Section 14.4. Today we explore properties of Galois extensions and their Galois groups. Proposition 1. Let K=F be a Galois extension, and F 0=F any extension. Then KF 0=F 0 is Galois with Galois group Gal(KF 0=F 0) ' Gal(K=K \ F 0): Proof. Since K=F is Galois, K is the splitting field of some separable polynomial f(x) 2 F [x]. This means that KF 0 is the splitting field of f(x) considered as a polynomial in F 0[x]. Hence KF 0=F 0 is Galois as well. Every element of Gal(KF 0=F 0) permutes roots of f(x) 2 F 0[x], and therefore preserves K. Then we have a restriction homomorphism 0 0 φ: Gal(KF =F ) −! Gal(K=F ); σ 7−! σjK : Clearly, this homomorphism is injective, since every σ 2 ker(φ) acts trivially on both F 0 and K, hence on KF 0. Now, let H = im(φ) ⊂ Gal(K=F ), and KH be the corresponding fixed subfield. Note that KH ⊃ K \ F 0, since every σ 2 Gal(KF 0=F 0) fixes F 0. On the other hand, KH F 0 is fixed by Gal(KF 0=F 0), hence KH F 0 = F 0, which implies KH ⊂ F 0 and finally KH = K \ F 0. 0 Therefore, H = Gal(K=K \ F ). Corollary 2. Suppose K=F is a Galois extension, and F 0=F is any finite extension. Then [K : F ][F 0 : F ] [KF 0 : F ] = : [K \ F 0 : F ] Proof. Indeed, by the above Proposition we have [KF 0 : F 0] = [K : K \ F 0]: Then [K : F ][F 0 : F ] [KF 0 : F ] = [KF 0 : F 0][F 0 : F ] = [K : K \ F 0][F 0 : F ] = : [K \ F 0 : F ] Proposition 3. Let K1=F and K2=F be Galois extensions. Then (1)( K1 \ K2)=F is Galois; (2) K1K2=F is Galois with the Galois group H = f(σ; τ) j σjK1\K2 = τjK1\K2 g ⊂ Gal(K1=F ) × Gal(K2=F ): Proof. Recall that K=F is Galois if and only if it is a finite, normal, and separable extension. Now, if both K1 and K2 are finite, normal, and separable, then so is their intersection. This proves (1). For part (2), if K1 and K2 are splitting fields for a pair of separable polynomials f1(x) and f2(x), then K1K2 is the splitting field for the square-free part of f(x) = f1(x)f2(x), hence is Galois as well. Now, the homomorphism φ: Gal(K1K2=F ) −! H; σ 7−! (σjK1 ; σjK2 ) is clearly injective. The order of H can be computed as j Gal(K1=F )jj Gal(K2=F )j jHj = j Gal(K1=F )jj Gal(K2=K1 \ K2)j = : j Gal(K1 \ K2=F )j The latter is equal to j Gal(K1K2=F )j by the previous Corollary, which completes the proof. 1 2 Corollary 4. If K1=F and K2=F are Galois extensions and K1 \ K2 = F , then Gal(K1K2=F ) ' Gal(K1=F ) × Gal(K2=F ): Conversely, if K=F is Galois with Galois group G ' G1 × G2 for a pair of subgroups G G G G G1;G2 ⊂ G, then K 1 \ K 2 = F and K = K 1 K 2 . Proof. The first part follows immediately from the previous Proposition. For the second, G G G G note that K 1 \K 2 is fixed by both G1 and G2 hence by G = G1×G2. Then K 1 \K 2 = F . G G G \G Similarly, K 1 K 2 = K 1 2 = K. Corollary 5. Let E=F be a finite separable extension. Then there exists a Galois extension K=F such that E ⊂ K, and for any other Galois extension L=F such that E ⊂ L we have K ⊂ L. Such K is called the Galois closure of E over F . Proof. Let α1; : : : ; αn be a basis of E over F , and p1(x); : : : ; pn(x) 2 F [x] be the corre- sponding minimal polynomials. Then, the composite of their splitting field contains E and is Galois over F . Now, we can set K to be the intersection of all Galois extensions of F containing E. Proposition 6. Let K=F be a finite extension. Then K = F (θ) if and only if there exists only finitely many subfields of K containing F . Proof. Let K = F (θ) and F ⊂ E ⊂ K. Let f(x) 2 F [x] and g(x) 2 E[x] be the minimal polynomials for θ. Let E0 be the field generated over F by the coefficients of g(x). On one hand, E0 ⊂ E, on the other, g(x) is still the minimal polynomial for θ over E0, which implies [K : E] = deg(g) = [K : E0]; hence E = E0. Therefore, every subfield E ⊂ K containing F is generated over F by the coefficients of monic factors of f(x) which shows that there are only finitely many such subfields E. Conversely, suppose there are only finitely many such subfields. If F is a finite field, and K its finite extension, we have already proved that K=F is simple, hence we can suppose that F is infinite. Now, K is finitely generated over F , and therefore it is enough to show that F (α; β) is simple for any α; β 2 K. Consider subfields F (α + cβ) with c 2 F . Since we only have finitely many such subfields, we can find a pair c 6= c0 such that F (α + cβ) = F (α + c0β). Then, α + c0β 2 F (α + cβ) from which it is easy to derive that F (α; β) = F (α + cβ). This completes the proof. Theorem 7. If K=F is finite and separable, then K=F is simple. Proof. Let L be the Galois closure of K over F . Then any subfield F ⊂ E ⊂ K corresponds to a subgroup of Gal(L=F ). But there are only finitely many subgroups of a Galois group, and by previous Proposition K=F is simple. Corollary 8. Any finite extension of a field of characteristic 0 is simple. As follows from the proof of the last Proposition, a primitive element for an extension can be obtained as a linear combination of the generators of the extension. In Galois case it is enough to require that the linear combination is not fixed by any non-trivial element of the Galois group. Example. p p p p (1) As we have seen before, 2 + 3 generates Q( 2; 3) over Q. 3 (2) Let Fp be the algebraic closure of the finite field Fp. Then the field of rational functions p p Fp(x; y) is not a simple extension of its subfield Fp(x ; y ). Indeed, we can see that p p 2 Fp(x; y): Fp(x ; y ) = p ; while p p Fp(x + cy): Fp(x ; y ) = p for any c 2 Fp, and there are infinitely many such subfields none of which are isomor- p p phic, since otherwise we would have Fp(x + cy) = Fp(x ; y )..
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