Let $(R,\mathfrak{m})$ be a regular local ring of dimension $d$. Let $P$ be a prime ideal of height $d-1$. I want to know if $P^2$ is always a $P$ primary ideal ie if $P/P^2$ is torsion free as $R/P$ module. Thanks.

Let me explain few of my observations. Assume that $R$ is a complete local ring. Then by Cohen's structure theorem $R$ is a power series ring over a DVR. In eqicharacteristic case it is actually a power series over a field. So can we prove the statement in the following special cases.

  1. $R=k[[X_1, X_2, \ldots, X_n]]$.
  2. $R=k[X_1, X_2, \ldots, X_n]_{(X_1, X_2, \ldots, X_n)}$.

Note that if $R/P$ is regular then $P$ is generated by a sequence. If $dimR=2$ then $P$ is principal . So in either cases $P^2$ is $P$ primary. In this problem we have to prove or disprove that $\mathfrak{m} \notin Ass R /P^2$ or $R/P^2$ is one dimensional cohen macaulay ring.


1 Answer 1


This is false. If $P$ is generated by a regular sequence then things should be ok.

Just some terminology that might be useful, since this sort of question is studied heavily in commutative algebra. The $P$-primary component of $P^n$ is called the $n$th symbolic power of $P$. In particular, there are numerous papers comparing the symbolic and ordinary powers of ideals, proving they are the same in some cases, and different in others, and finding effective bounds for when one ideal is contained in another.

Ok, so here's the counter-example that I had in mind:

$$R := QQ[a,b,c]_{\langle a,b,c\rangle},\;\; P := \langle b^2 - ac, a^2b + ac - c^2, a^3 + ab - bc \rangle$$

This is an ideal of a curve that crosses over itself three times at the origin (but is smooth elsewhere).

It is not hard to check that the square of this ideal is not primary, and I've done that by hand in the past (if I recall correctly). However, in this case I just checked it in Macaulay2 by running the commands

  1. $R = QQ[a,b,c]$
  2. $P := \text{ideal}(b^2 - a*c, a^2*b + a*c - c^2, a^3 + a*b - b*c)$
  3. $\text{primaryDecomposition}(P^2)$

I didn't bother localizing, but localizing at the origin won't impact the primary decomposition at the origin.

  • $\begingroup$ I have checked it with Singular. Answer is perfect. Thanks. $\endgroup$ Mar 9, 2012 at 5:30

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