(Krull-Akizuki) If $R$ is a one-dimensional
Noetherian domain with quotient field $K$, and $L$ is a finite
extension field of $K$, then any subring $S$ of $L$ that contains $R$
is Noetherian and of dimension $\leq 1$ , and has only finitely many
ideals containing a given nonzero ideal of R. In particular, the
integral closure of $R$ in $L$ is Noetherian.
The main part of the proof is about showing that if $0\neq J\subset S$ be an ideal, by proving that $J/aS$ for some $0\neq a\in J\cap R$, is an $R$-module of finite length. And then, it concludes that the rest of the assertions follows.
I am not quite sure about the dimension $\leq 1$ part of the theorem. I think it might be related to the incomparability of distinct prime ideals, but this property holds when $R\subset S$ is an integral extension. However, in the situation above, $S$ is obviously an algebraic extension over $R$. I think this might be somehow true even though, or is there any other points that I missed ?
Best Answer
It is also proved that $\ell_R(S/aS)<\infty$, where $a\in R$, $a\ne0$. If $P$ is a non-zero prime ideal of $S$, then $P\cap R\ne(0)$ and $S/P$ is a quotient of $S/aS$ for some $a\in P\cap R$, $a\ne 0$. But $S/aS$ is an artinian $R$-module, hence an artinian ring. It follows that $S/P$ is an artinian domain, hence a field. We showed that $P$ is maximal, and thus $\dim S\le1$.