Height of associated prime ideal is zero

Question

Let $P\in\operatorname{Ass}(R)$ in a Noetherian ring $R$, and assume the local ring $R_P$ is a domain. I want to prove that the height of $P$ is zero.

I know that in a Noetherian ring, each ideal (here the zero ideal) contains a power of its radical. Also, I tried the well-known fact that in a primary decomposition $I=Q_1∩...∩Q_n$ in a Noetherian ring $R$, the prime ideals $P_i=√Q_i$ are exactly those among $\{√(I:x)|x\in R\}$, but could reach nowhere. Incidentally, for a local ring $(S, m)$ we have the Krull dimension equal to zero if and only if $m$ is nilpotent. So, perhaps this might be applied to $S=R_P$.

I would thank anybody giving suggestions!

Answer 1

We have $\mathfrak p=\operatorname{Ann}(a)$ for some $a\in R$, $a\ne0$. Then $ar=0$ for all $r\in\mathfrak p$. Since $R_{\mathfrak p}$ is a domain this leads to $\frac a1=\frac 01$ or $\frac r1=\frac01$.
If $\frac a1=\frac01$ then there is $s\in R-\mathfrak p$ such that $sa=0$, so $s\in\mathfrak p$, a contradiction.
Hence $\frac r1=\frac01$ for all $r\in\mathfrak p$, that is, $\mathfrak pR_{\mathfrak p}=0$ and this shows us that $R_{\mathfrak p}$ is a field, so the height of $\mathfrak p$ is zero.