Let $R$ be a commutative Noetherian ring and let $p$ be a maximal ideal of $R$.
The localization of $R$ at $p$, $R_p$, is a local ring with unique maximal ideal $pR_p$.
Now let $I$ be an arbitrary (= not necessarily prime) ideal of $R$ such that $I \subseteq p$.
Assume that $IR_p=pR_P$.
Question: Is it true that $I=p$?
Of course, if we knew that $I$ is a prime ideal of $R$, then by the known result
concerning the one-one correspondence between prime ideals of $R_p$ and prime ideals of $R$ contained in $p$, we would have obtained that $I=M$.
However, here $I$ is not known to be a prime ideal of $R$.
Relevant questions: 1, 2, 3; the second answer in reference 3 quotes Theore 5.32 from "Steps in commutative agebra" by Sharp,
and it seems that I need some version of Theorem 5.30, just without the primality assumption (still with contraction and extension of ideals).
Remark:
$R=\mathbb{Z}$, $p=0$ is not a counterexample.
Edit: What if $R$ is an integral domain?
Thank you very much!
Edit 2: Now asked this more general question.
Best Answer
No, this is not true, even if $R$ is an integral domain. Consider the case when $R$ is the polynomial ring $F[x]$ over your favorite field $F$, and let $\mathfrak{p}=\langle x\rangle$ and $I=\langle x(x-1)\rangle$. Then $\mathfrak{p}$ is a maximal ideal and $I$ is a strict subset of $\mathfrak{p}$, but, in the localized ring $R_\mathfrak{p}$, the element $(x-1)\big/1$ is a unit, and we hence have $$I_{\mathfrak{p}}=\left\langle x(x-1)\big/1\right\rangle\ni \left(x(x-1)\big/1\right)\left(1\big/(x-1)\right)=x\big/1,$$ whence $I_\mathfrak{p}\supseteq\mathfrak{p}_{\mathfrak{p}}=\left\langle x\big/1\right\rangle$.