Integral closure of Gaussian Integers

Question

I am considering $\mathbb{Z}[i]\subset\mathbb{Q}(i)$

Now I have a short note here that says that there are elements of $\mathbb{Q}(i)$ which are not integral over $\mathbb{Z}[i]$ 'because $\mathbb{Q}(i)$ is a field but $\mathbb{Z}[i]$ is not'.

Can somebody please explain to me why the statement follows from this fact?

Answer 1

Given any integral ring extension $B\subseteq A$, then Krull dimensions of $A$ and $B$ coincide. In particular, if $B$ is a field then $A$ as well. In our case here, $B=\mathbb{Q}(i)$ is a field, but $A=\mathbb{Z}[i]$ is not, hence $B\subseteq A$ is not integral. So there must be an element of $B$ which is not integral over $A$.

Answer 2

$\mathbf Z[i]$ is the integral closure of $\mathbf Z$ in the extension field $\mathbf Q\subset \mathbf Q(i)$, hence it is integrally closed, i. e. the only elements of $\mathbf Q(i)$ which are integral over $\mathbf Z[i]$ (root of a monic polynomial with coefficients in $\mathbf Z[i]$) belong to $\mathbf Z[i]$.

Answer 3

Let $D$ be a domain and $Q$ be its field of fractions. Then $Q$ is integral over $D$ iff $D$ is a field.

One direction is trivial. For the other, suppose $Q$ is integral over $D$. Take $u \in D, u\ne0$. Then $1/u \in Q$ is integral over $D$ and so satisfies $(1/u)^n+a_{n-1} (1/u)^{n-1} + \cdots + a_1 (1/u) +a_0=0$ with $a_i \in D$. This implies that $1+a_{n-1} u + \cdots + a_1 u^{n-1} +a_0 u^{n}=0$ and so $1$ is a multiple of $u$ in $D$. This means that $u$ is unit in $D$. Therefore $D$ is a field.