Why is the Discriminant always an integer? In my Galois theory notes, I read that we define the $discriminant$ of a degree $n$ polynomial $f$ with roots $\alpha_1, ... , \alpha_n $ as 
$$\Delta^2 = (-1)^{n(n-1)/2}\prod\limits_{i\ne j}(\alpha_i-\alpha_j)$$
And when I look at the discriminants of various low degree polynomials (over the integers) I notice that they are always integers:
$n = 2:{\displaystyle b^{2}-4ac\,}$
$n=3:{\displaystyle b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd\,}$
${n=4:\displaystyle {\begin{aligned}{}&256a^{3}e^{3}-192a^{2}bde^{2}-128a^{2}c^{2}e^{2}+144a^{2}cd^{2}e\\&{}-27a^{2}d^{4}+144ab^{2}ce^{2}-6ab^{2}d^{2}e-80abc^{2}de\\&{}+18abcd^{3}+16ac^{4}e-4ac^{3}d^{2}-27b^{4}e^{2}+18b^{3}cde\\&{}-4b^{3}d^{3}-4b^{2}c^{3}e+b^{2}c^{2}d^{2}\,\end{aligned}}}$
Etc. So the question I want to ask is; is this always the case? If so, how do we prove this? And more importantly, if we look at arbitrary polynomials in the ring $K[t]$ for some base field $K$, is the discriminant always in the base field $K$?
Many thanks.
 A: Another definition of the discriminant of a polynomial
$
f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0
$
is
$$
\Delta = \frac{(-1)^\frac{n(n-1)}{2}}{a_n}\operatorname{resultant}(f,f')
$$
The resultant is a determinant with entries in $K$ and so is in $K$.
For a monic polynomial with coefficients in a ring $R$, the discriminant is in $R$.
A: Take any element of the Galois group $g \in \operatorname{Gal}(f)$ and consider how it acts on the discriminant: $$g(\Delta^2) = (-1)^{n(n-1)/2} \prod_{i \neq j}(g(\alpha_i) - g(\alpha_j))$$
Now $g$ permutes the roots $\alpha_i$ in some order, but since the product runs over all $i \neq j$, we still get all $(\alpha_i - \alpha_j)$ turning up in the product. We deduce that $g(\Delta^2) = \Delta^2$ and so, by Galois theory, $\Delta^2 \in \mathbb{Q}$.
Your definition of the discriminant assumes that $f$ is monic. In this case, the roots $\alpha_i$ are algebraic integers, and therefore so is $\Delta^2$.
Putting this together, $\Delta^2$ is both a rational number and an algebraic integer, hence a rational integer $\Delta^2 \in \mathbb{Z}$.
