Let $charK \neq 2$, $L/K$ be the splitting field of a separable polynomial $f \in K[x]$. Further let $a_1,...,a_n \in L$ be the roots of $f$.
How does the Galoisgroup $Gal(L/K)$ act on $c:=\prod_{1 \leq i < j \leq n}(a_i-a_j)$?
$Fix Gal(L/K) \cap A_n =K(c)$
My attempt:
About 1):
As far as I know for $\sigma \in Gal(L/K)$ we have that $\sigma$ permutates the roots $a_1,...,a_n$. Now (by intuition) I would say that
$\sigma(\prod_{1 \leq i < j \leq n}(a_i-a_j))=\prod_{1 \leq i < j \leq n}(\sigma(a_i)-\sigma(a_j))=sgn(\sigma) \prod_{1 \leq i < j \leq n}(a_i-a_j)$
But I am not really sure how to answer the question in Nr.1.
About 2):
I know that $Gal(L/K)$ should be a subgroup of the symmetric group $S_n$. $A_n$ denotes the even permutations, and $A_n \leq S_n $. If $\sigma \ in Fix Gal(L/K) \cap A_n$, then we have that $\sigma(c)=c$
My only approach would be to use the teorem about the primitive element:
Let L be s field extension of the form $L=K(a,c_1,...,c_n)$ such that $a$ is algebraic over $K$ and $c_1,...,c_n$ are separable over $K$, then there exists a primitive element $b \in L $ such that $L=K(b)$.
But since this theorem only says something about the existence, I don't think that it would be useful.