1
$\begingroup$

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 $\mathfrak pA_\mathfrak p$.

$\endgroup$
  • $\begingroup$ Or could have added in the hypothesis that $A$ is noetherian. $\endgroup$ – user26857 Dec 29 '14 at 22:09
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Whoops, Konstantin posted a similar answer two minutes ago. Unicity of natural answers? $\endgroup$ – Georges Elencwajg Dec 28 '14 at 22:20
  • $\begingroup$ Looks like it! Seems hard to think of other examples actually. $\endgroup$ – Konstantin Ardakov Dec 28 '14 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.