7

Just as the title says.

Let $R$ be a Noetherian integral domain, let $K$ be its field of fractions, let $L$ be a finite extension of $K$, and let $S$ be the integral closure of $R$ in $L$. Must $S$ be Noetherian, or do I need some additional assumptions on $R$?

EDIT: I meant to assume that $R$ itself is integrally closed in $K$ to start with. Does that change things?

user26857
  • 53,190
calearner
  • 305

2 Answers2

6

This is true if $\dim R = 1$, and is known as the Krull-Akizuki theorem. In fact, it is commonly stated with the stronger conclusion that any subring $S \subseteq L$ containing $R$ is Noetherian. If $\dim R = 2$, it is still true that $\overline{R}^L$ is Noetherian, although there may be subrings of $L$ that are not. In dimension $3$ though, Nagata has given examples of $3$-dimensional Noetherian domains whose integral closures are not Noetherian.

If $R$ is a Nagata ring (e.g. an excellent ring, and thus virtually any geometric ring), then the desired conclusion holds.

zcn
  • 16,235
  • Thank you! But I knew what you said above already. I stated the problem wrong to begin with because I forgot to add the crucial assumption that $R$ starts out being integrally closed in its fraction field. What happens then? – calearner Feb 08 '14 at 02:33
  • The Wikipedia link above has an example of a DVR which is not a Japanese ring, so your conclusion still would not hold for such a ring. Apparently the fraction field being perfect is enough though (so e.g. in characteristic zero) – zcn Feb 08 '14 at 05:19
  • I don't quite see that yet. $R$ not being Japanese just means that $S$ doesn't have to be finitely generated $R$-module (or equivalently, it doesn't have to be f.g. as an $R$-algebra)... but that doesn't answer the question of whether $S$ is Noetherian or not, does it? – calearner Feb 08 '14 at 22:54
  • 1
    You're right: $S$ is not Noetherian as an $R$-module, but it could still be Noetherian as an $S$-module, i.e. as a ring (which is what I assume you meant in your question). I'm afraid I don't have a counterexample to your edit. Just a remark: If $L/K$ is separable, then $S$ is module-finite over $R$, hence Noetherian (see e.g. this question) – zcn Feb 09 '14 at 02:09
  • Nagata's counterexample is cited in Serre's book "Local Algebra" as M. Nagata, Note on integral closure of Noetherian domains, Mem. Univ. Kyoto, 28, 1953, 121-124. – Watson Jan 14 '21 at 09:48
  • The dimension 2 case is due to Nagata (https://www.math.purdue.edu/~iswanso/book/SwansonHuneke.pdf, 4.10.6). – Watson Jan 14 '21 at 10:32
0

If $L/K$ is a finite separable extension, then $S$ is Noetherian.

Lemma. $S$ is a finitely generated $R$-module.

Theorem. $S$ is a Noetherian ring.

Proof. Suppose $I \subset S$ is an ideal. By the lemma above, $S$ is a finitely generated $R$-module. Since $R$ is Noetherian, $I$ is a finitely generated $R$-module. Of course $I$ is also a finitely generated $S$-module, i.e., a finitely generated ideal of $S$, hence $S$ is Noetherian.

Zephyr
  • 121