the ideal contains some power of the unique prime ideal containing it Let $R$ be a Noetherian ring, and $I$ an ideal such that there exists a unique prime ideal $\mathfrak{p}$ containing $I$. Show that $I$ contains some power of $\mathfrak{p}.$
In this question, I couldn't use the uniqueness of the prime ideal, and being that $R$ is a Noetherian ring. Could anyone give some hints?
 A: Alex's solution is the most straightforward. One can also observe that $R/I$ is a local ring of dimension $0$, since $I$ is contained in a unique prime (which, upon modding out by $I$, becomes the maximal ideal of $R/I$). And a Noetherian ring of dimension $0$ is an Artin ring, in which every non-trivial ideal is nilpotent. Thus $\mathfrak p/I$ is nilpotent in $R/I$.
A: Consider the quotient ring $R/I$, and let $\pi \colon R \to R/I$ be the canonical quotient homomorphism. The nilradical of $R/I$ is the image under $\pi$ of the intersection of all prime ideals of $R$ containing $I$. Since there is a unique prime ideal $\mathfrak{p}$ containing $I$, the nilradical of $R/I$ is $\pi(\mathfrak{p})$. Now, this means that every element of $\pi(\mathfrak{p})$ is nilpotent in $R/I$. Since $R$ is noetherian, $R/I$ is also noetherian, so $\pi(\mathfrak{p})$ is finitely generated by elements $x_{1}, \ldots, x_{k}$. Can you wrap it up from here to show that every element of $\pi(\mathfrak{p})$ vanishes when raised to a sufficiently large power $N \in \mathbb{N}$, thus showing that $\mathfrak{p}^{N} \subset I$? 
A: No need to assume "there exists a unique prime ideal p containing I" but assume "there exists a unique *minimal* prime ideal p containing I". 
1- We need  "$R$ is a Noetherian", to have $(\sqrt I)^n\subseteq I;$ for some $n\in \mathbb N$. (This is proposition 7.14 of Atiyah-Macdonald).
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2- By "uniqueness of the prime ideal containing $I$", we have $\sqrt I = p$. (Proposition 1.14 of Atiyah-Macdonald says The radical of an ideal is intersection of prime ideals containing it. ).  
