Galois Group Problem

  • #1
I am trying to solve question 7 on this problem sheet. This is my progress so far:

Using the Leibniz formula for the determinant of a matrix I deduce that

[tex] \det(\Omega) = \sum_{\tau \in S_n} sign(\tau) \prod_{i = 1}^n \sigma_i (w_{\tau(i)}) [/tex]


Hence [tex] P + N = \sum_{\tau \in S_n}\prod_{i = 1}^n \sigma_i (w_{\tau(i)}) [/tex]

and [tex] PN = (\sum_{\tau \in A_n}\prod_{i = 1}^n \sigma_i (w_{\tau(i)}))\cdot (\sum_{\tau \in S_n \setminus A_n}\prod_{i = 1}^n \sigma_i (w_{\tau(i)})) [/tex]

[tex] = \sum_{\tau \in S_n; \pi \in S_n \setminus A_n}\prod_{i = 1}^n \sigma_i (w_{\tau(i)}w_{\pi(i)}) [/tex]

Having not done a formal course in Galois Theory, I'm a bit unsure about the significance of M here. I realise that Aut(M/Q) can only be called Gal(M/Q) if M:Q is normal (implied by it's being a splitting field) and separable (automatic in C).

So Gal(M/Q) is a the group of automorphisms of M that fix Q and thus due to the Galois Correspondence, the subgroup of Gal(M/Q) of elements that fix K corresponds uniquely to K and vice versa. I don't quite see how this helps here though.

Since

[tex] Aut(K/Q) = \{\sigma_1 , \ldots , \sigma_n \} [/tex]

I see that applying any of these to the above expressions would leave them invariant, but what I don't see is why they are invariant under Gal(M/Q). After all M seems to be arbitrary.
 

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Answers and Replies

  • #2
gel
533
5
[tex] Aut(K/Q) = \{\sigma_1 , \ldots , \sigma_n \} [/tex]
No, I don't think that's right. I think that [itex]\sigma_i[/itex] are the embeddings of K into C. Note that, as M is normal and contains K, they are also embeddings into M and are permuted by applying the elements of the Galois group of M/Q. The question can then be approached by considering what happens to P and N under even and odd permutations.
 
  • #3
OK thanks. Yeah the \sigma 's are the embeddings of K. The only thing about what you say that doesn't make sense to me is the bit about them being permuted by the elements of the Galois group of M/Q. The elements of this group permute the Q-automorphisms of M by definition, but I don't see how this implies that they permute the embeddings of K.
 
  • #4
gel
533
5
each embedding sigma is a homorphism K->C, whose image will in fact be in M, so can be regarded as a homomorphism K->M. An element g of the Galois group of M/Q is a homomorphism M->M. So, by composition, [itex]g\sigma:K->M[/itex] which is just another embedding of K into M (and therefore, into C).
 
  • #5
Thanks - excellently explained!
 

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