Solving Galois Group Problem Q7: Step-by-Step Guide

[Hello Kitty]

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

\det(\Omega) = \sum_{\tau \in S_n} sign(\tau) \prod_{i = 1}^n \sigma_i (w_{\tau(i)})Hence P + N = \sum_{\tau \in S_n}\prod_{i = 1}^n \sigma_i (w_{\tau(i)})

and 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)}))

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

Having not done a formal course in Galois Theory, I'm a bit unsure about the significance of M here. I realize 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

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

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.

 

[gel]

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

No, I don't think that's right. I think that \sigma_i 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.

 

[Hello Kitty]

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.

 

[gel]

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, g\sigma:K->M which is just another embedding of K into M (and therefore, into C).

 

[Hello Kitty]

Thanks - excellently explained!