3 Examples Involving Small Galois Groups

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1 3 Examples Involving Small Galois Groups In this section we work out some of the details from recent homework problems and some additional results for Galois groups of orders 4 and 8. Throughout this section we will assume all fields have characteristic other than 2. Proposition 3.1. Let L/F be an extension of degree 4. Then L = F (α) where α is a root of a polynomial of the form x 4 + ax 2 + b if and only if L contains a quadratic intermediate field. Proof. If L = F (α) where α is a root of the given equation, then α = ± a± a 2 4b, and hence F (α 2 )=F( a 2 2 4b). This must be a quadratic extension of F,orelse F (α) would be of degree 2 over F, not 4. Conversely, if L/F is an extension of degree 4 having a quadratic intermediate field K = F ( d), then L is a quadratic extension of K, and so necessarily L = K( β),β K. Then β = r + s d K, and so L = F (α) where α = β. Then α is arootofx 4 2rx 2 +r 2 s 2 d, which is a polynomial of the required form. Proposition 3.2. Let f(x) =x 4 + ax 2 + b F [x]. Let±α, ±β be the roots of f(x) where α = a+ a 2 4b. Then f(x) is irreducible over F if and only if 2 α 2,α± β/ F. Proof. We have f(x) =(x α)(x + α)(x β)(x + β). If f(x) isreducible, then it factors as two quadratics (possibly further reducible). Since (x + α) must occur in one of the quadratic factors, there are three possibilities for the factor in which it occurs: (x + α)(x α) =x 2 α 2, (x + α)(x + β) = x 2 +(α + β)x + αβ, and (x + α)(x β) =x 2 +(α β)x αβ. Since each of these factors would need to be in F [x], we see that if f(x) factors, then at least one of α 2,α+ β,α β must be in F. Conversely, if α 2 F, then f(x) =(x α 2 )(x β 2 ) gives a factorization in F [x] (and so β 2 f(x) also). If α ± β F, then (α 2 + β 2 ± 2αβ) F. Now α 2 + β 2 = a, so2αβ F, and so αβ F. Then x 2 +(α ± β)x ± αβ F [x], so again f(x) isreducible over F. Theorem 3.3. Let f(x) be as above, and assume f is irreducible over F. Let G = Gal(f(x)) be the Galois group of f over F. Then 1. G = V, the Klein four-group, if and only if b F 2,ifand only if αβ F. 14

2 2. G = C, the cyclic group of order 4, ifand only if b(a 2 4b) is a square in F,ifand only if F (αβ) =F (α 2 ). 3. G = D 8, the dihedral group of order 8, ifand only if b and b(a 2 4b) are not squares in F,ifand only if αβ / F (α 2 ). Proof. First notice that the Galois closure of F (α) isf (α, β), but β 2 F (α), so [F (α, β) :F ]=4or8. Now D = 4αβ(α 2 β 2 ) 2 = 4 b[(α 2 + β 2 ) 2 4α 2 β 2 ]= 4 b(a 2 4b), so D F if and only if b = αβ F. But D F = G A4, and the only possible group of order 4 inside A 4 is V. If, however, D/ F, then G A 4, and G = V. This proves (1). Now assume b/ F. Note that a 2 4b / F or else α 2 F, contradicting the irreducibility of f(x). Then since G = V, the only possibilities for G are C or D 8.Wehave G = C G has a unique quadratic intermediate field. This happens if and only if F ( b)=f (αβ) =F (α 2 )=F( a 2 4b), if and only if b(a 2 4b) F 2. This leaves G = D 8 under the conditions not yet considered, specifically when b = αβ / F (α 2 ). The following proposition concerning the cyclic group of order 4 C 4 can be derived fairly easily from our analysis above. Proposition 3.4. The group C 4 is realizable as a Galois group over the field F if and only if F contains an element ɛ which is not a square in F, but is a sum of two squares in F.AnyC 4 -extension can be realized via a tower of quadratic extensions F F ( ɛ) F ( a + b ɛ), where a 2 ɛb 2 = ɛz 2 for some z F. The minimal polynomial of this extension is x 4 2ax 2 +a 2 ɛb 2. If f(x) =x 4 + cx 2 + d is an irreducible polynomial over F, then Gal(f) = C 4 if and only if d/ F 2, but d(c 2 4d) F 2. The field F ( ɛ) can be embedded in a cyclic extension of degree 4 precisely when ɛ/ F 2 and ɛ is a sum of two squares in F. A field which fails to have C 4 as a Galois group, i.e. a field in which every sum of squares is in fact a square, is called a pythagorean field. Pythagorean fields are of considerable interest in quadratic form theory. This next result also belongs to the collection of Galois-theoretic folklore. We provide an outline of the proof here. Theorem 3.5. Let a, b F, independent mod F 2. There exists a Galois extension L/F with Gal(L/F ) = D, F F ( a, b) L, and with 15

3 Gal(L/F ( ab)) = C, ifand only if the equation ax 2 +by 2 = z 2 has a nontrivial solution over F.Any such extension can be written as F ( a, b, z + x a), where z 2 ax 2 = by 2 for some y F. Proof. Suppose first that L is a D-extension. Then there are five intermediate subfields of codimension 2 in L. Let K be one of them such that b/ K. Then K = F ( a, t) for some t F ( a), and it follows that L = K( b)= F ( a, b, t). Since L is a C-extension of F ( ab) containing F ( ab, b)as its quadratic intermediate field, we see from the results above that t = z+y b for some z,y F ( ab) where the norm of t over F ( ab),z 2 by 2,isequal to bx 2 for some x F ( ab). Since also t F ( a), it follows that this is also the norm of t over F, and so N F ( a)/f (t) b F ( ab) 2 F = bf 2 af 2.If N(t) af 2,wewould have F ( a, t)being a C-extension of F, which is impossible as C is not a quotient of D. ThusN(t) bf 2, and so z 2 ax 2 = by 2 has a nontrivial solution in F,asclaimed. Conversely, suppose t F ( a)issuch that N(t) =by 2 y F. Let L = F ( a, b, t). One can verify that L is indeed Galois over F, that Gal(L/F ( ab) = C, and letting τ be agenerator of Gal(L/F ( ab)) and σ the generator of Gal(L/F ( a, t)), it can be checked that σ, τ together generate Gal(L/F ), and that σ, τ = D. The group Q has received extensive attention in the literature. The complete description of when a given biquadratic extension F ( a, b)off can be embedded in an extension with Galois group Q was first given by Witt(1936). We remark that there is a misprint in the description of the extensions realizing Q in Witt s paper: the summand 1 was left out of the expression under the square root. Some of the terminology in the statement of the following theorem will be unfamiliar. I will explain, as time allows, in class. Theorem 3.6. (Witt, 1936) Let F be afield with a, b F, independent mod F 2. The following conditions are equivalent. 1. There exists a Galois extension L of F, with Gal(L/F ) = Q, and such that F ( a, b) is the unique biquadratic intermediate field between F and L. 2. (a, b)(a, a)(b, b) =1 Br(F ). 3. The quadratic form ax bx ab x2 3 is equivalent to the quadratic form y 2 1 +y 2 2 +y 2 3 under a change of variables x i = p ij y j, with det(p ij )=1. 16

4 The extensions realizing Q are given by the fields ( ( ) ) 1 F r 1+p 11 a + p22 b + p33, ab with r F. The obstruction to the embedding of the biquadratic extension F ( a, b) in a degree 8 extension with quaternion Galois group is the class of (a, b)(ab, 1) = ( a, b)( 1, 1) Br 2 (F ). Inparticular, if F ( a) embeds in a Q-extension, then a is a sum of three squares in F. However, the converse is not true. We conclude with a couple of results concerning Galois closures of Galois p-extensions of Galois extensions. Proposition 3.7. Let p a prime, F a field. Let K/F be a p-extension (i.e. a Galois extension with Gal(K/F) a p-group). Let L be a p-extension of K. Then the Galois closure of L over F is a p-extension of F. Proof. Since L/K is separable and K/F is separable, it follows that L/F is separable. It is also finite, and so L = F (θ) for some θ L. Let f(x) be the minimal polynomial for θ over F, and let the roots be θ = θ 1,θ 2,...,θ n. Then if M is the Galois closure of L/F, we have M = F (θ 1,...,θ n ) = K(θ 1,...,θ n ). Since K/F is Galois, it must be that K F (θ i ) i. Wewill be done if we can show that each L i = F (θ i )isap-extension of K, for then taking successive composites and using Proposition 1.3, we see M = L 1 L 2 L k is Galois over K and is of degree a power of p over K, hence is a p-extension of F. Thus the key is to show that L i is Galois over K. Since each L i = L over F, the degree over K is necessarily a power of p. Let G = Gal(M/F), N = Gal(M/K), H i = Gal(M/L i ). Then H = H 1 N and N G, since L/K and K/F are Galois. We need to see that H i N. Inother words, we need to see that for all τ N,τH i τ 1 = H i. We know there exists σ G such that L i = σ(l), and thus H i = σhσ 1. Then τh i τ 1 = τσh(τσ) 1. Since N G, wehave Nσ = σn σ G, sothere exists τ N such that τσ = σ τ. Then τh i τ 1 = σ τh τ 1 σ 1 = σhσ 1 = H i, since H N. Thus H i N as desired. Perhaps a simpler proof that L i is Galois was provided in a student s solution: Let g(x) bethe minimal polynomial for θ over K. Then since L i = σ(l) for some σ G, the minimal polynomial for θ i = σ(θ) over K is σ(g(x)). 17

5 (Notice that σ(g) K[x] because K/F is Galois and hence σ(k) = K.) If θ = α 1,α 2,...,α r L are the roots of g(x), then σ(α 1 ),σ(α 2 ),...,σ(α r ) σ(l) =L i are the roots of σ(g). Then L i is the splitting field for σ(g) over K, just as L is the splitting field for g, and hence L i is Galois over K. Proposition 3.8. Let K/F be agalois extension,and let L = K( β) be a quadratic extension of K, where β K \ K 2. Then L/F is Galois if and only if βσ(β) K 2 σ Gal(K/F). Proof. Let f(x) = n i=1 a ix i be the minimal polynomial for β over F,son = [F (β) :F ]. Then β satisfies g(x) = a i x 2i, and by degree considerations this must be the minimal polynomial for β over F. The roots of f(x) are {σ(β) σ Gal(K/F)} and there are n distinct such roots. Then clearly {± σ(β) σ Gal(K/F)} gives the 2n distinct roots of g(x). We have L/F Galois if and only if L contains all the roots of g(x), i.e. if and only if K( σ(β)) = K( β) for all σ Gal(K/F). But this happens if and only if β σ(β) K, ifand only if βσ(β) K 2. 18