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Let $(R,m)$ be a $1$-dimensional noetherian local domain and $S$ its integral closure. Clearly $S$ is $1$-dimensional noetherian semi-local domain. Is $mS=J(S)$, where $J(S)$ is the Jacobson radical of $S$?

user26857
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alinaghi
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  • I don't think so. Take $(R,m)=(p\Bbb Z^{(p)}, (p^2))$. The normalization of $(R,m)$ is $S=(\Bbb Z^{(p)},(p))$, but $m\Bbb Z^{(p)}=(p^2)\ne (p)=J(S)$ – Adam Hughes Jul 22 '14 at 17:23
  • @AdamHughes What is $\mathbb Z^{(p)}$? – user26857 Jul 22 '14 at 20:14
  • @AdamHughes: Is your counterexample a ring without 1? – zcn Jul 23 '14 at 01:10
  • @user26857 the $p$ local integers, i.e. ${a/b: a,b\in\Bbb Z,,\gcd (b,p)=1}$ – Adam Hughes Jul 23 '14 at 01:13
  • @zcn yes. Taken after the general orders in a number field, $k$, (i.e. integral ideals) and their integral closure/normalization $\mathcal{O}_k$. Since the op wanted a local version, I localized a specific example. Neukirch uses them as typical Nötherian domains of dimension $1$. – Adam Hughes Jul 23 '14 at 01:14
  • @AdamHughes: That ring is typically written $\mathbb{Z}_{(p)}$, as it is a localization of $\mathbb{Z}$ at the prime ideal $(p)$. Although what you point out is true, giving a ring without $1$ doesn't seem particularly relevant, especially when talking about domains – zcn Jul 23 '14 at 01:18
  • @zcn possibly, but I'm pretty confident that depends upon your personal context. In any case, it works for those who work with the CRing definition which does not require a $1$. – Adam Hughes Jul 23 '14 at 01:24

1 Answers1

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One can use the following proposition to find a counterexample:

Proposition: Let $R$ be a Noetherian complete local domain. If $R \subseteq S \subseteq \text{Quot}(R)$ and $S$ is integral over $R$, then $S$ is local.

Proof: $R \subseteq S$ is module-finite (since under the assumptions, the integral closure $\overline{R}$ is module-finite over $R$, see e.g. Swanson-Huneke, Theorem 4.3.4). The result then follows e.g. from Corollary 7.6 in Eisenbud.

Thus, it suffices to give an example of a $1$-dim Noetherian complete local domain $(R,m)$ such that for $S = \overline{R}^{\text{Quot}(R)}$, $mS$ is not a prime ideal. One such ring is $R = k[[t^3,t^4,t^5]]$. Then $S = k[[t]]$, and $mS = (t^3,t^4,t^5)S \ne tS = J(S)$.

zcn
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  • The proposition really holds for an integral extension? I've tried to use Nagata's theorem (about the integral closure of a noetherian complete local domain in a finite field extension of its field of fractions) to prove it, and I think I need $R\subset S$ finite. On the other side, you don't need the proposition in all its generality; in fact, you use Nagata's theorem – user26857 Jul 23 '14 at 16:34
  • @user26857: I did not think too much about proving it, but from what I recall, this is true. As you say though, this does not affect the example at hand, which is a finite extension – zcn Jul 23 '14 at 22:07
  • @user26857: After amending to make a slightly more modest statement, I have added a proof (by reference) – zcn Aug 29 '14 at 17:58