# How to compute the integral closure of $\Bbb{Z}$ in $\mathbb Q(\sqrt{p})$?

We have the definition of integral closure that all the integral elements of A in B. Could we just compute the integral closure of certain A in B. I am considering such a problem that given a prime p, what is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[x]/(x^n–p)$.

After some trials, I find the answer maybe $\mathbb{Z}[x]/(x^n–p)$. Since $\mathbb{Z}\subset\mathbb{Z}[x]/(x^n–p)$, the integral closure of $\mathbb{Z}$ should be the subset of the closure of $\mathbb{Z}[x]/(x^n–p)$. But $\mathbb{Z}[x]/(x^n–p)$ is generated by $1, s, s^2, \dots, s^{n–1}$ which are integral over $\mathbb{Z}$. $s$ is the image of x modulo (x^n–p). Therefore $\mathbb{Z}[x]/(x^n–p)$ is integral over $\mathbb{Z}$.

Now I want to prove that $\mathbb{Z}[x]/(x^n–p)$ is normal.
Assume that $u\in\mathbb{Q}[x]/(x^n–p)$ is integral over $\mathbb{Z}[x]/(x^n–p)$ with a monic poly $f$. Since $\exists d \in\mathbb{Z}$ and $e\in \mathbb{Z}[x]/(x^n–p)$ s.t du–e=0. Then f can be divided by dx-e. Assume that $f=(dx-e)g$, where $g$ is a poly with leading coefficient h in $\mathbb{Z}[x]/(x^n–p)$.
Then we get $hd=1$. Since $d$ is a integer, $h$ has to be a integer too. This implies that $d=1$ or $–1$. Both sides implies $u\in \mathbb{Z}[x]/(x^n–p)$. So we got the integral closure of $\mathbb{Z}$ contains $\mathbb{Z}[x]/(x^n–p)$ which is exactly the closure of itself. Q.E.D

Is there any mistakes in the proof?