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I try to solve some question about galois theory.. Let $f(x)$ be irredicuble polynomial over $\mathbb{Q}$ and $f(x)=(x-\alpha_1)...(x-\alpha_n)$ where $\alpha_i \in \mathbb{C}$

splitting field is $K=\mathbb{Q}(\alpha_1,...,\alpha_n)$

define discriminant as follows

$D=\prod_{i<j}(\alpha_{i}-\alpha_{j})^2$

Firs questionn is that

Show that $K/Q$ is galois extension and $\operatorname{Gal}(K/Q)$ is subgroup of $S_n$.

I don't understand relation between discriminant and this question. I showed directly, since it is normal and seperable, it must be Galois. and there is n roots so that subgroup of $S_n$ Is it true?

And I can not show the other questions which are

Show that $D\in\mathbb{Q}$.

If $\sqrt{D}\in \mathbb{Q}$ then $\operatorname{Gal}(K/Q)$ is subgroup of $A_n$.

(I did something about this one, I assumed $\sqrt{D}\in \mathbb{Q}$. $K=\mathbb{Q}(\alpha_1,..., \sqrt{D})$ since $\sqrt{D}\in \mathbb{Q}$, that is a subgroup of $S_n$ But how can I show that it is subgroup of alternating group?)

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  • $\begingroup$ I don't think the first part is meant to have anything to do with the discriminant. $\endgroup$
    – Hoot
    Nov 1 '15 at 16:09
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to show that $D\in Q$ show that it is fixed by elements of the Galois group.

$\sqrt{D}=\prod(\alpha_i-\alpha_j)$, let $s$ in the Galois group $s(\sqrt{D})=sign(s)\sqrt{D}$, thus $s(\sqrt{D})=\sqrt{D}$ iff $sign(s)=1$.

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  • $\begingroup$ I don't understand.. what does it means? $\endgroup$ Nov 1 '15 at 17:39

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