$R=\frac{\mathbb{Z}[x]}{}$ is integrally closed or not?

Question

Suppose $$R_4=\frac{\mathbb{Z}[x]}{<x^2+4>}$$ Then $R$ is not integrally closed because the element $\frac{\overline{x}}{\overline{2}}$ outside $R$, lying in the fraction field of $R$ is integral because it satisfies the equation $F(Y)=Y^2+1$

In general Suppose $$R_{k^2}=\frac{\mathbb{Z}[x]}{<x^2+k^2>}$$ Then $R$ is not integrally closed because the element $\frac{\overline{x}}{\overline{k}}$ outside $R$, lying in the fraction field of $R$ is integral because it satisfies the equation $F(Y)=Y^2+1$

So naturally I was thinking what if we consider the ideal $x^2+n$ where $n$ is not a perfect square.

Question

Suppose $$R_3=\frac{\mathbb{Z}[x]}{<x^2+3>}$$ Then is $R$ integrally closed ?

I assumed that $\frac{\overline{f}}{\overline{g}}$ in the fraction field of $R_3$ be integral, and it satisfies some monic polynomial over $R$, then multiplied by $g^n$ to get an equation in $\mathbb{Z}[x]$ but cannot proceed further.

So for me, it was easier to show that it is not integrally closed, rather than showing integrally closed.

Answer 1

In general, if $K=\mathbb{Q}(\sqrt{d})$ with $d$ a square-free integer such that $d\equiv2$ or $3\mod 4$, then its ring of integers is given by $\mathcal{O}=\mathbb{Z}[\sqrt{d}]$. If $d\equiv 1\mod 4$ then you need a bigger ring: $\mathcal{O}=\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$. Rings of integers are normal.

Your ring is isomorphic to $\mathbb{Z}[\sqrt{-3}]$ which has fraction field $\mathbb{Q}(\sqrt{-3})$. Therefore your ring is not normal. The element of its fraction field $$\omega=\frac{-1+\sqrt{-3}}{2}$$ satisfies $\omega^2+\omega+1=0$. The integral closure of $\mathbb{Z}[\sqrt{-3}]$ is the ring of Eisenstein integers.