# Galois extension and discriminant

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

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?)

• I don't think the first part is meant to have anything to do with the discriminant.
– Hoot
Nov 1, 2015 at 16:09

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 $sign(s)=1$.