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Question is :

Show that if $R$ is a Noetherian ring, and $I$ is an ideal such that there exists a unique prime ideal $\mathfrak{p}$ containing $I$, then $I$ contains some power of $\mathfrak{p}$.

Hint given is:

First show that if an element $x$ is contained in every prime ideal of a ring then $x^n=0$ for some $n$. Do this by considering the family of all ideals such that no power of $x$ lies in them. Next, consider the ring $R/I$.

Consider the ideal $P/I$ in ring $R/I$. This is the only prime ideal in $R/I$.

Let $x\in P$ then $x+I\in P/I$ the only prime ideal in $R/I$.

Which then imply that $(x+I)^n=0+I$ i.e., $x^n\in I$. This says, fixing an element $x\in P$ we have $x^n\in I$.

This $n$ may not work for all elements in $P$.

I am not very sure how to use that $R$ is noethrian to conlcude that there is unique $n$ that works for all elements of $P$.

Please give only hints.

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Hint: Since $R$ is Noetherian $P$ is finitely generated.Let $p_1,..,p_l$ generators of $P$, $p_i^{n_i}\in I$ implies that $x^{n_1+..+n_l}\in I$ for every $x\in P$.

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