# If $t \in R$ generates a prime ideal in $R$ then $R$ is normal Iff $R_t$ is normal and $R_{(t)}$ is DVR.

Let $R$ be a Noetherian ring. If $t\in R$ generates a prime ideal in $R$ show that $R$ is normal iff $R_t$ is normal and $R_{(t)}$ is a DVR.

I'm trying to solve the converse of this exercise. Let $\mathfrak p$ be a prime ideal in $R$. If $t \not \in \mathfrak p$ then I have proved that $R_{\mathfrak p}$ is an integrally closed domain. If $t \in \mathfrak p$ and $\mathrm{ht}(\mathfrak p)=1$ then $R_{\mathfrak p}$ is an integrally closed domain. But I am having trouble in proving this in the case $t \in \mathfrak p$ and $\mathrm{ht}(\mathfrak p) \geq2$.

Any hints/ideas.

• @user26857, My intuition was Serre's normality criterion. I proved that if $\mathfrak p$ be a prime ideal in $R$, such that $\mathrm{ht}(\mathfrak p)= 0 or 1$ then $R_{\mathfrak p}$ is regular. And if $\mathfrak p$ be a prime ideal in $R$ of $\mathrm{ht} \geq 2$ and $t \not \in \mathfrak p$ then $\mathrm{depth} R_{\mathfrak p}\geq 2$. But i'm having trouble in proving it in the case if $t \in \mathfrak p$. – user270331 Jan 17 '18 at 5:18

If $\mathfrak p$ is a prime ideal in $R$ such that $t \in \mathfrak p$ and $\mathrm{ht}(\mathfrak p) \geq2$ then $(t)R_{\mathfrak p}$ is a prime ideal in $R_{\mathfrak p}$ of height $1$. Hence $R_{\mathfrak p}$ is a domain and $t/1$ is $R_{\mathfrak p}$-regular (by Exercise 16.10 of Matsumura's Commutative Ring Theory). Moreover, $R_{\mathfrak p}/(t)R_{\mathfrak p}$ is a domain of dimension $\geq1$. From which we can conclude that $\operatorname{depth}R_{\mathfrak p}\geq2$.