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I had asked this question on math.stackexchange.com, but I have not received a response. Hoping to get some help here.

I am trying to prove this result and I am stuck at one step. Let $(R,m)$ be a Noetherian local ring. Let $P$ be a prime ideal. Define the n-th symbolic power of $P$ as $P^{(n)}={{r\in R: \exists y \in R\setminus P}}$ s.t. $yr \in P^n$.

I want to show $P^{(n)}=P^n:m^{\infty}$.

Let $r\in P^n:m^{\infty}$. Then, $\exists t$ s.t. $rm^t\in P^n$. Now, $m^t \subsetneq P^n$, else by taking radicals, we have $m=P$. Let $y\in m^t \setminus P$. Then, $yr\in P^n$, so that $r\in P^{(n)}$.

For the converse, let $r\in P^{(n)}$. Suppose, $y\in R\setminus P$ s.t. $ry\in P^n$. If $y\in R\setminus m$, then $y$ is a unit and $r\in P^n\subseteq P^{(n)}$.

I am not sure how to handle the case, where $y\in m\setminus P$. Is it true $yR$ is $m$-primary? This would imply $yR$ contains a power of $m$ and thus completing the proof.

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  • $\begingroup$ No, $yR$ is not $m$-primary, since it has height at most one by the Principal Ideal Theorem. I'm pretty sure the result you're trying to prove is false, but can't see why immediately. $\endgroup$
    – Graham Leuschke
    Commented Jul 6, 2011 at 18:00
  • $\begingroup$ Something similar is true, replacing the maximal ideal by the intersection of the non-minimal associated primes of $P^n$. See Examples 4.3 of these lecture notes of Herzog: cel.archives-ouvertes.fr/docs/00/37/46/20/PDF/CIMPAHerzog.pdf. In particular I think what you want is true in dimension $\leq 2$. $\endgroup$
    – Graham Leuschke
    Commented Jul 6, 2011 at 18:13
  • $\begingroup$ @Graham: I guess I am missing some hypothesis then. In section 4, in the following paper, this is the formula used for calculating symbolic powers.people.reed.edu/~iswanson/topology.pdf Is the statement true if $R$ is regular, local? $\endgroup$
    – Brittany Murphy
    Commented Jul 6, 2011 at 21:13
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    $\begingroup$ This is false for the following reason. $P^n:m^{\infty}/P^n$ is a module of finite length, since all elements are annihilated by a power of the maximal ideal. In general $P^{(n)}/P^n$ is not. To make an example, take an example of some prime ideal in a power series ring in $k$ variables such that $P^n\neq P^{(n)}$. Then consider the same inside a power series ring with one more variable. $\endgroup$
    – Mohan
    Commented Jul 6, 2011 at 22:40
  • $\begingroup$ @Mohan: I am not sure I understand your comment. The author uses this formula in section 4 of this paper that I referenced above people.reed.edu/~iswanson/topology.pdf in a power series ring with 3 variables. $\endgroup$
    – Brittany Murphy
    Commented Jul 7, 2011 at 0:11

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