An ideal of a Noetherian ring with one associated prime

Question

Suppose that for a Noetherian ring $R$, we have an ideal $I$ such that $\operatorname{Ass}(R/I)$ has only one element, $P$.

Then $\forall r \in R, p \in P, \; rp \in I \Rightarrow Rp = (p) \subset I$

$\Rightarrow RP = P \subset I$

But clearly $\forall r \in R, a \in I, \; ra \in I \Rightarrow I \subset P$

Thus we must have $I = P$ is a prime ideal.

Is my reasoning for this correct?

Answer 1

I think you're using the fact that there exists an $x \in R$ with $P = \text{ann}(x + I)$ to conclude that $rp \in I$ in the first line. However, note that $\text{Ass}(R/I)$ being a singleton does not mean that for all $x \in R$, $P = \text{ann}(x + I)$. It only means that there exists one such $x$.

Answer 2

A counter-example:

Let $p$ be a prime number and $n>1$ an integer. $\mathbf Z/p^n\mathbf Z$ is a local noetherian ring, with a single, non-zero, prime ideal: $\;p\mathbf Z/p^n\mathbf Z$, which is therefore associated to the $0$ ideal. Yet the maximal ideal is not $0$.