# Liu, exercise 2.1.4: Minimal prime ideals and nilpotents

In the book "Algebraic Geometry and Arithmetic Curves" Liu wrote in errata that there is a mistake in this problem:

Let $A$ be a commutative ring with unit. (a) Let $\mathfrak p$ be a minimal prime ideal of $A$. Show that $\mathfrak pA_{\mathfrak p}$ is nilpotent.

What would be a counterexample of this exercise stated as above as Liu wrote in http://www.math.u-bordeaux1.fr/~qliu/Book/errata-third-b1.pdf that one should consider every element of $\mathfrak pA_\mathfrak p$ rather that $\mathfrak pA_\mathfrak p$.

• Or could have added in the hypothesis that $A$ is noetherian. – user26857 Dec 29 '14 at 22:09

Let $B$ be the localisation of a polynomial algebra $k[x_1,x_2,\ldots]$ in infinitely many variables over a field $k$ at the ideal $(x_1,x_2,\ldots)$. This is a local, non-Noetherian ring. Now let $A = B / (x_1, x_2^2, x_3^3, x_4^4, \ldots)$. This ring is still local with maximal ideal $\mathfrak{m}$ say, and every element of $\mathfrak{m}$ is nilpotent. Now, if $\mathfrak{p}$ is a prime ideal of $A$ and $x \in \mathfrak{m}$ then $x^n = 0\in\mathfrak{p}$ for some $n$ which forces $x \in \mathfrak{p}$. Thus $\mathfrak{m}$ is contained in every prime ideal of $A$. So $\mathfrak{m}$ is a minimal prime ideal in $A$, and $\mathfrak{m} A_{\mathfrak{m}} = \mathfrak{m}$ since $A$ is already local. But $\mathfrak{m}$ is not nilpotent by construction.
Take $A=\mathbb Q[X_1,X_2,\cdots,X_n,\cdots]/\langle X_1,X_2^2,\cdots,X_n^n,\cdots\rangle=\mathbb Q[x_1,x_2,\cdots,x_n,\cdots]$ and $\mathfrak p=\langle x_1,x_2,\cdots,x_n,\cdots\rangle$ .
Since $\mathfrak p$ is actually the only prime ideal of $A$, it is minimal and maximal so that we have $A_\mathfrak p=A$ and $\mathfrak pA_\mathfrak p=\mathfrak p$.
It now suffices to notice that all elements of $\mathfrak p$ are nilpotent but that no power $\mathfrak p^N$ of $\mathfrak p$ is zero since $x^N_{N+1} \neq 0$.
Summing up, $\mathfrak p$ is nil but not nilpotent.