Integral Closure in $\mathbb{Q}(\alpha)$

Question

Let $\alpha $ be a root of $f(x) = x^3 -x +3$, $K=\mathbb{Q}(\alpha)$ I have to prove $\mathcal{O} _K = \mathbb{Z}[\alpha]$ for the integral closure $\mathcal{O} _K$. I have shown that the disciminant of $\mathbb{Z}[\alpha]$ is -239 and should now be able to deduce $\mathcal{O} _K = \mathbb{Z}[\alpha]$ (using that 239 is a prime number I guess) ....and for this I need some help.

I also have a second similar problem: Let $\alpha $ be a root of $f(x) = x^3 -x^2 -2x-2$, $K=\mathbb{Q}(\alpha)$. Again I have to show $\mathcal{O} _K = \mathbb{Z}[\alpha]$ (using Stickelbergers criterion: the discriminant of any $n$-tuple in $\mathcal{O} _K$ is $0$ or $1$ mod $4$) What I was able to do here was to show that the discriminant of $\mathbb{Z}[\alpha]$ is -220, but again I don't know what to next

Thanks in advance!

Edit: the second discriminant is -152...sign flaw

Answer 1

The following fact should help (try proving this -- it's not too hard):

If $A \subset B$ are free abelian groups, then $\operatorname{disc} A = [B:A]^2 \operatorname{disc} B$.

It follows that if $\operatorname{disc} \mathbb Z[\alpha]$ is squarefree that $\mathbb Z[\alpha] = \mathcal O_K$.

This fact should also help on your second problem.