 #1
 25
 0
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.
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.
Attachments

37.1 KB Views: 138
Last edited: