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

This preview shows page 18 - 21 out of 69 pages.

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
Image of page 18
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
Image of page 19
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 | .
Image of page 20
Image of page 21

You've reached the end of your free preview.

Want to read all 69 pages?

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

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes