Proof that a domain is normal 
Given the domain $\mathbb Z[\alpha]$, with $\alpha^2+2\alpha+4=0$, prove that it is normal.

We basically saw two facts about normal domains: one is that any UFD is normal (but $\mathbb Z[\alpha]$ isn't a UFD); the other is the following theorem: let $A$ be a normal domain and $K$ its field of fractions. Let $L$ be an extension of $K$. Then an element of $L$ is integral over $A$ iff the coefficients of its minimal polynomial over $K$ are all in $A$.
However the only way that I see to apply this theorem is with $A=\mathbb Z$, $K=\mathbb Q$ and $L=\mathbb Q[\alpha]$, and it doesn't seem to be useful, since  $\mathbb Z[\alpha]$ is not involved. Can you give me a hint on how to start?
 A: $\mathbb Z[\alpha]\simeq\mathbb Z[X]/(X^2+2X+4)$. But $\mathbb Z[X]/(X^2+2X+4)\simeq\mathbb Z[X]/(X^2+3)$ (send $X$ to $X-1$) which is at its turn isomorphic to $\mathbb Z[\sqrt{-3}]$. This is not integrally closed since $\frac{1+\sqrt{-3}}{2}$ is integral over $\mathbb Z[\sqrt{-3}]$ and does not belong to it.
In fact, the integral closure of $\mathbb Z[\sqrt{-3}]$ is $\mathbb Z[\frac{1+\sqrt{-3}}{2}]$. This gives us a hint about the question you mentioned in a comment: the least $n\ge 2$ such that $\mathbb Z[\alpha,n^{-1}]$ is integrally closed is $2$. In order to show that $\mathbb Z[\sqrt{-3},\frac12]$ is integrally closed let $x\in\mathbb Q(\sqrt{-3})$, the field of fractions of all three rings we mentioned in this paragraph, and suppose that $x$ is integral over $\mathbb Z[\sqrt{-3},\frac12]$. Then $2^kx$ is integral over $\mathbb Z[\sqrt{-3}]$ for some non-negative integer $k$. Since it is integral over $\mathbb Z[\frac{1+\sqrt{-3}}{2}]$ which is integrally closed, we have $2^kx\in\mathbb Z[\frac{1+\sqrt{-3}}{2}]$, and this implies that $x\in\mathbb Z[\sqrt{-3},\frac12]$.
