Subring of number field is Noetherian

Question

I've been trying to prove that every subring $R$ of a number field $K$ is a Noetherian ring. I'm aware that of a proof when $K = \mathbb{Q}$ using the fact that every subring of $\mathbb{Q}$ is a localization of $\mathbb{Z}$. However, this fact crucially uses that $\mathbb{Z}$ is a PID, hence only answers my question in a special case. By Hilbert's basis theorem, I've been able to prove the result when $R$ is a finitely generated $\mathbb{Z}$-algebra. Beyond that, I'm stuck.

Answer 1

First, I claim that for any prime $p\in\mathbb{Z}$, the quotient $R/pR$ is finite dimensional as an $\mathbb{F}_p$-vector space. Indeed, suppose $x_1,\dots,x_n\in R$ have images in $R/pR$ that are linearly independent over $\mathbb{F}_p$. I claim that $x_1,\dots,x_n$ are then linearly independent over $\mathbb{Z}$ (and hence over $\mathbb{Q}$), so $n\leq [K:\mathbb{Q}]$. Indeed, suppose that $\sum a_kx_k=0$ for some integers $a_k$ which are not all $0$. Let $m$ be maximal such that $p^m\mid a_k$ for all $k$. Then $\sum \frac{a_k}{p^m}x_k=0$ as well, and some $\frac{a_k}{p^m}$ is not divisible by $p$. We can then take this equation mod $p$ to get a linear dependence between the images of the $x_k$ in $R/pR$, which is a contradiction.

So $R/pR$ is finite for any prime $p\in\mathbb{Z}$. It follows that $p^mR/p^{m+1}R$ is finite for all $m$ (since it is a cyclic $R/pR$-module) and thus $R/p^mR$ is finite by induction on $m$. By the Chinese remainder theorem it follows that $R/nR$ is finite for any nonzero integer $n$.

Now note that any nonzero $a\in R$ is a divisor (in $R$) of some nonzero integer. Indeed, if $f(x)\in\mathbb{Z}[x]$ is the minimal polynomial of $a$ over $\mathbb{Q}$, then $f(a)=0$ says that $a$ divides the constant term of $f$. It follows that $R/aR$ is finite since it is a quotient of $R/nR$ for some nonzero integer $n$.

So, for any nonzero $a\in R$, there are only finitely many ideals of $R$ that contain $aR$. That is, every nonzero ideal of $R$ is contained in only finitely many other ideals. It follows that $R$ is Noetherian.