V. Fields and Galois Theory

Transcription

1 Math 201C - Alebra Erin Pearse V.2. The Fundamental Theorem. V. Fields and Galois Theory 4. What is the Galois roup of F = Q( 2, 3, 5) over Q? Since F is enerated over Q by {1, 2, 3, 5}, we need to determine all possible imaes σ( 2), σ( 3), σ( 5) 1 for σ AutQF. Now 2 has irreducible polynomial f(x) = x 2 2 Q[x] 3 has irreducible polynomial (x) = x 2 3 Q[x] 5 has irreducible polynomial h(x) = x 2 5 Q[x] Since σ must take roots of f to roots of f (etc.), we must have ( ) σ 2 = ± ( ) 2 σ 3 = ± ( ) 3 σ 5 = ± 5 Since we have three independent binary choices, AutQF = Z 2 Z 2 Z 2. To verify this formally, note that AutQF = 8 and every element u of AutQF has order u Let f K(x) with f of K by K(x). / K and f, relatively prime in K[x] and consider the extension a) x is alebraic over K( f ) and [K(x) : K( f )] = max (de f, de ). Since f/ is transcendental over K, we write f/ = z and consider z as an indeterminate. Let the polynomial ϕ K(z)[y] be defined by ϕ (y) = z (y) f (y) so that x is clearly a root of ϕ. First we show that it is primitive and irreducible over K[z][y] = K[z, y], and then use III ϕ is irreducible in K[z][y]. Suppose ϕ were reducible: but then < ϕ (y) = (z 1 (y) f 1 (y)) h (where f 1 h = f and 1 h = ), f, are relatively prime. ϕ is primitive. Suppose not: then for ϕ (y) = (zb 0 a 0 ) + (zb 1 a 1 ) y (zb n a n ) y n there would be r K such that zb i a i = b i (z r), which implies a i = rb i, i. But this would mean f = r K, < hypothesis. Now that ϕ is a primitive irreducible polynomial, we know ϕ is irreducible over K(z)[y] by III Hence, ϕ is the irreducible polynomial of x and x is thus alebraic over K( f ). By examination of ϕ, it is also clear that [K(x) : K( f )] = de ϕ = max (de f, de ). 1 Since σ fixes K, we know that we must have σ(1) = 1.

2 2 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse b) If E K is an intermediate field of the extension K(x) : K, then [K(x) : E] <. E K, so pick u E\K. Then u E implies that u is of the form f(x), for some (x) f, K[x]\K. x is alebraic over K(u) by part (a), because E K = f / K and we can choose f, coprime. Let h be the irreducible polynomial of x over K(u). Then we have a tower of fields K (x) E K (h) K and a correspondin product of dimensions [K (x) : K (u)] = [K (x) : E] [E : K (u)]. Since [K (x) : K (u)] = de h <, this clearly implies [K(x) : E] <. c) x f ϕ(x) induces a homomorphism such that σ : ϕ (f/)/ψ (f/). ψ(x) σ Aut K F max (de f, de ) = 1. To see that σ induces the iven homomorphism, use the proof of 2.2. Note that σ is a field homomorphism, so it is certainly injective by III.2.21(iv). Now, [ ( )] f max (de f, de ) = 1 K (x) : K ( ) = 1 by (a) K (x) = K = σ (K (x)) d) Aut K K(x) = {x ax+b cx+d First note that the map ϕ(x) f and thus fixes K. Then f σ is surjective.. a, b, c, d K and ad bc 0} = GL 2 (K). ψ(x) ϕ(f/) ψ(f/) σ Aut K K(x) σ GL 2 (K) follows immediately from (c) and the fact that For a K\{0}, b K, the maps σ a : K(x) K(x) by σ a : f(x) (x) σ is invertible ad bc 0. f(ax) (ax) and are clearly K-automorphisms of K(x). discussed previously is just evaluation at τ b : K(x) K(x) by τ b : f(x) (x) f(x+b) (x+b)

3 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse 3 9. a) If K is infinite, then K(x) is Galois over K. If K(x) is not Galois over K, then K F (Aut K K(x)), so we must have K E = F (Aut K K(x)). Then [K(x) : E] < by 6(b), which implies that [ AutE K(x) : Aut K(x) K(x) ] = [Aut E K(x) : 1] = Aut E K(x) [K(x) : E] by Lemma 2.8 < By hypothesis, K =, so Aut K K(x) = { ax+b cx+d} = by 6(d). But then Lemma 2.6(iv) ives which then implies < above. Aut K K(x) = Aut E K(x) Aut E K(x) = b) If K is finite, then K(x) is not Galois over K. First note that, as above, Aut K K(x) = ax+b cx+d by 6(d), so K = n < = Aut K K(x) n 4 < By hypothesis, [Aut K K(x) : 1 K ] <, so [F (1 K ) : F (Aut K K(x))] [Aut K K(x) : 1 K ] by Lemma 2.9. If K(x) were Galois, this would imply < x is transcendental. [K(x) : K] Aut K K(x) n 4 <

4 4 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse 11. In the extension of Q by Q(x), show that the intermediate field Q(x 2 ) is closed, but Q(x 3 ) is not. a) Q(x 2 ) is closed u Q(x)\Q(x 2 ), σ AutQ(x 2 )Q(x) such that σ(u) u. Note that Q(x) is alebraic over Q(x 2 ) with basis {1, x}, because x is a root of f(y) = y 2 x 2 Q(x 2 )[y]. Then pick u Q(x)\Q(x 2 ), so u = a + bx where b 0. Let σ be defined by σ(1) = 1, σ(x) = x. Now σ(u) = a bx u, but for any v Q(x 2 ), define v = n i=0 a ix 2i so that σ(v) = n σ (a i) σ (x) 2i i=0 Thus, σ fixes Q(x 2 ) but not u. = n a i ( x) 2i i=0 = n a ix 2i i=0 = v b) To see that Q(x 3 ) is not closed, we will exhibit u Q(x)\Q(x 3 ) which is fixed by any σ AutQ(x 3 )Q(x). Let u = x + x 3 so that σ(u) = σ(x) + x 3 because σ fixes x 3 Q(x 3 ). We note that Q(x) is alebraic over Q(x 3 ) with basis {1, x, x 2 }, because x is a root of f(y) = y 3 x 3 Q(x 3 )[y]. So σ is completely determined by its action on {1, x}. We know that σ(1) = 1 as above (because σ must fix Q(x 3 )), but what is σ(x)? σ(x) must be a root of f(y), but f(y) has only one root: x. Thus, σ(x) = x = σ(u) = x + x 3 = u, for any σ which fixes Q(x 3 ). Thus, u F ( AutQ(x 3 )Q(x) ) \Q(x 3 ) = F ( AutQ(x 3 )Q(x) ) Q(x 3 ), i.e., Q(x 3 ) is not closed.

5 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse Let F : K be a finite-dimensional Galois extension with intermediate fields L, M. a) Aut LM F = Aut L F Aut M F LM is the smallest field containin L and M means that L LM, M LM and L, M E = LM E We proceed to show the equality by a double-inclusion arument. L LM = Aut LM F Aut L F, and M LM = Aut LM F Aut M F, so clearly Aut LM F Aut L F Aut M F. Makin extensive use of the fact that F : K is f.d.., Aut L F Aut M F Aut L F, Aut M F = F (Aut L F ), F (Aut M F ) F (Aut L F Aut M F ) = L, M F (Aut L F Aut M F ) Lemma 2.10 = LM F (Aut L F Aut M F ) by first line = Aut F(AutL F Aut M F )F Aut LM F by Fun. Thm. = Aut L F Aut M F Aut LM F Lemma 2.6(iii) b) Aut L T M F = Aut L F Aut M F L M L, M = Aut L F, Aut M F Aut L M F = Aut L F Aut M F Aut L M F, because the join is the smallest roup containin both Aut L F and Aut M F. F (Aut L M F ) = L M by f.d.., so Aut L F, Aut M F Aut L F Aut M F def of = F (Aut L F Aut M F ) F (Aut L F, Aut M F ) by Fun. Thm. = F (Aut L F Aut M F ) L, M by f.d.. = F (Aut L F Aut M F ) L M c) What conclusion can be drawn if Aut L F Aut M F = 1? In this case, Aut L F Aut M F = Aut LM F = 1 by above = F (Aut LM F = F ) by Fun. Thm. = LM = F by f.d..

6 6 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse 15. a) F : K is a finite-dimensional Galois extension with intermediate field E =!L where L is the smallest field such that E L F and L is Galois over K. Let {L i } be the set of Galois extensions of K which contain E. Since [F : K] <, there are a finite number of them. Now define L = n i=1 L i so that L is the smallest extension of K containin E. To see that L : K is Galois, consider the correspondin subroups Aut Li F. We have Aut Li F Aut K F for each L i, by the Fun. Thm., and n Aut L F = Aut Li F = Aut Li F i=1 follows by #14(b). Now we have n Aut Li F Aut K F i=1 by exercise I.5.3(a) (with a brief induction). Then Aut L F G = L : K is Galois by the Fun. Thm. aain.

7 June 7, 2002 Math 201C - Alebra - V.2 Erin Pearse 7 b) Furthermore, Aut L F = σ σ (Aut EF ) σ 1 where σ runs over all of Aut K F. To see Aut L F σ σ (Aut EF ) σ 1, Aut L F = σ (Aut L F ) σ 1 σ σ (Aut E F ) σ 1 σ σ (Aut E F ) σ 1 σ Aut K F Aut L F Aut K F Aut L F Aut E F Now to see σ σ (Aut EF ) σ 1 Aut L F. We know that Aut L F = Aut Li F, so it suffices to show that σ σ (Aut EF ) σ 1 Aut Li F for some i, i.e., that σ σ (Aut EF ) σ 1 is normal in Aut K F and contained in Aut E F. To see σ σ (Aut EF ) σ 1 Aut K F, pick τ Aut K F. Then ) τ ( σ σ (Aut EF ) σ 1 τ 1 = τ ( σ (Aut E F ) σ 1) τ 1 σ = σ τσ (Aut EF ) (τσ) 1 = ρ ρ (Aut EF ) ρ 1 = σ σ (Aut EF ) σ 1. To see why the last steps hold, consider that the intersection runs over all permutations σ Aut K F, but it doesn t matter in which order. Puttin στ = ρ amount to shufflin the indices and reorderin them in their oriinal state. Now σ σ (Aut EF ) σ 1 is invariant under conjuation and hence normal, which means that σ σ (Aut EF ) σ 1 = Aut Lj F for some j. Thus, σ σ (Aut EF ) σ 1 Aut Li F = Aut Li F = Aut L F.