0
$\begingroup$

Let $(R,\mathfrak m)$ be a local domain of dimension $1$. Let $\overline R$ be the integral closure of $R$ in the field of fractions $Q(R)$. If $R$ is henselian, then is $\overline R$ also a local ring ?

Thoughts: According to https://stacks.math.columbia.edu/tag/04GH , if $\overline R$ were moreover module finite over $R$, then $\overline R$ would be a product of local rings, and then $\overline R$ being an integral domain would imply it is local. However, $\overline R$ would be module finite over $R$ if and only if the $\mathfrak m$-adic completion $\widehat R$ is reduced (see Theorem 4.6 of the book https://bookstore.ams.org/surv-181), so I do not see if this approach will work (unless of course we assume $R$ is complete, then we are done).

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Yes, without restriction on dimension. See https://stacks.math.columbia.edu/tag/0BQ0.

Improve this answer
$\endgroup$
2
  • $\begingroup$ To follow up; if $R$ is a henselian domain, then by stacks.math.columbia.edu/tag/0C24, we have that $\overline{R}$ is a local ring since the number of minimal primes of $R^h$ is the same as the number of maximal ideals of $\overline{R}$, correct? $\endgroup$
    – walkar
    Commented Oct 31, 2023 at 12:47
  • $\begingroup$ @walkar: Yes, indeed. $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 31, 2023 at 13:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .