So \u03c3\u03b1 j \u03b1 i for some i Since \u03c3 is onetoone it follows that \u03c3 permutes the set \u03b1

# So σα j α i for some i since σ is onetoone it

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So σα j = α i for some i . Since σ is one–to–one, it follows that σ permutes the set { α 1 , α 2 , · · · , α n } . As these generate K/k , we have σ ( K ) = K . Definition 3.3 An extension K/k which is both separable and normal is said to be Galois . The group G of k –automorphisms of K is called the Galois group of K/k and is written Gal( K/k ). Theorem 2.5 implies that | G | = | K : k | . Associated with any subgroup H of G is the subfield of K given by Fix( H ) = { x K : σx = x σ H } . Associated with any intermediate field L (i.e. k L K ) is a subgroup Gal( K/L ) = { σ G : σ | L = id L } . Theorem 3.4 (The Fundamental Theorem of Galois Theory) Let K/k be a finite Galois extension with Galois group G . 1. The maps I : H Fix( H ) and J : L Gal( K/L ) between subgroups of G and fields intermediate between k and K are inverse bijections. 2. The maps I and J induce a correspondence between normal subgroups of G and normal extensions of k (contained in K ). If L/k is normal, with k L K , then Gal( L/k ) G/ ( L ). 3. I and J are lattice–preserving isomorphisms. This means (a) H K iff I ( H ) I ( K ). (b) I ( H K ) = I ( H ) I ( K ). (c) I H, K = I ( H ) I ( K ). Proof 17 1. First observe that clearly H ( J I )( H ) and L ( I J )( L ) for each H and L . So | ( J I )( H ) | ≥ | H | , and [ K : L ] [ K : ( I J )( L )]. Theorem 2.5 implies that | J ( L ) | = [ K : L ]. Theorem 2.4 implies that | H | ≤ [ K : I ( H )]. To establish the reverse inequality, we use the Primitive Element Theorem. Let α be a primitive element for K/k . Consider the polynomial f ( x ) = σ H ( x - σα ) . If τ H , the map σ στ permutes the elements of H . Hence this map preserves f . So f has coefficients in I ( H ) = Fix( H ), and thus [ K : I ( H )] ≤ | H | , as required. So | H | = [ K : I ( H )]. Hence we have | ( J I )( H ) | = [ K : I ( H )] = | H | and [ K : ( J I )( L )] = | J ( L ) | = [ K : L ], so J I and I J are both identity maps, as required. 2. Suppose first that H G , x Fix( H ), and τ G . For each σ H , we have στ ( x ) = τ ( τ - 1 στ )( x ) = τ ( x ), so τ ( x ) Fix( H ). Hence τ (Fix( H )) = Fix( H ) for each τ ( G ), and so Fix( H ) /k is a normal extension. Conversely, suppose that L/k is normal and τ Gal( K/L ). If l L and σ G , then σ = m L . But σ - 1 τσ ( l ) = σ - 1 τ ( m ) = σ - 1 ( m ) = l. Hence σ - 1 τσ Gal( K/l ), and so Gal( K/L ) G . The map σ σ | L induces the required isomorphism. 3. Exercise, e.g. I ( H ) I ( K ) = { x | σx = x for σ H K } = { x | σx = x for σ = H, K } = I H, K , etc. Example Let K denote the splitting field of x 4 - 2 over Q . Determine Gal( K/ Q ). Find all the subfields of K and all inclusion relations between them, identifying which ones are normal over Q . Let α R + satisfy α 4 = 2, and let i C satisfy i 2 = - 1. Since x 4 - 2 = ( x - α )( x + α )( x - )( x + ) , we have K = Q ( α, i ). Eisenstein implies that x 4 - 2 is irreducible over Q , so [ Q ( α ) : Q ] = 4. Moreover, i R Q ( α ), so [ Q ( α, i ) : Q ( α )] = 2, and hence [ K : Q ] = 8. Let X = { α, - α, iα, - } . Then G = Gal( K/ Q ) permutes X , and | G | = 8. Let S be a square with vertices labelled by the elements of X , and let D 8 be the group of rigid motions of S : α - α - 18 If σ G , then σ ( - α ) = σα , so σ ( - α ) and σα are “opposite” one another. Thus G D 8 , and so G = D 8 , as | G | = | D 8 | .  #### You've reached the end of your free preview.

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• Spring '08
• Morrison,D
• Algebra, Galois theory, Root of unity

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