I was recently reading about integral ring extensions. One of the first examples given is that $\mathbb{Z}$ is integrally closed in its quotient field $\mathbb{Q}$. Another is that $\mathbb{Z}[\sqrt{5}]$ is not integrally closed in $\mathbb{Q}(\sqrt{5})$ since for example $(1+\sqrt{5})/2\in\mathbb{Q}[\sqrt{5}]$ is integral over $\mathbb{Z}$ as a root of $X^2-X-1$, but $(1+\sqrt{5})/2\notin\mathbb{Z}[\sqrt{5}]$.
Now I'm curious, can we find what are all integers $n$ such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed (equal to its integral closure in its quotient field)?
One thing I do know is that unique factorization domains are integrally closed, so I think rings like $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{-2}]$, $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{3}]$ are integrally closed, as they are Euclidean domains, and thus are UFDs. But can we say what all integers $n$ are such that $\mathbb{Z}[\sqrt{n}]$ is integrally closed? Thanks!
Best Answer
Let's assume that $n$ is squarefree. I think it would be best to explore this without using knowledge of the answer. You want to find the integers in $\mathbf Q(\sqrt n)$. These are elements which satisfy a quadratic with integer coefficients. Every element looks like $a + b\sqrt{n}$ for some $a, b \in \mathbf Q$.
Recall that the conjugate of this element will be $a - b\sqrt n$. So it certainly satisfies \[ (X - a - b\sqrt n)(X - a + b\sqrt n) = X^2 - 2aX + a^2 - nb^2. \] You need the coefficients $-2a$ and $a^2 - nb^2$ to lie in $\mathbf Z$. (You might recognize these as the trace and norm.) If this is the case, then as $(2a)^2 - n(2b)^2$ is an integer (divisible by $4$) and $n$ is squarefree, it follows that $2b \in \mathbf Z$ as well.
At this point, it makes sense to work mod $4$: we know that this last expression vanishes there, and the only squares are $0$ and $1$. For example, if $n \equiv 2$ then $(2a)^2$ and $(2b)^2$ must both be congruent to $0$, and it follows that $a$ and $b$ are in $\mathbf Z$.