Integral closure of 1-dimensional noetherian local domains

Question

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 $S$?

Answer 1

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 $mS = (t^3,t^4,t^5)S \ne tS = J(S)$.