# Symbolic powers in regular local rings

I am having trouble understanding one of the results in the following paper http://arxiv.org/PS_cache/math/pdf/0104/0104175v1.pdf

In proposition 3.1, the author says

Let $(R,\frak{m})$ be a regular local ring, $p,q$ prime ideals s.t. $rad(p+q)= \frak{m}$ and $dim(R/p)+dim(R/q)=dim(R)$. If $R/p$ is regular then $p^{(m)}\cap q^{(n)}\subseteq p^{(m)}\frak{m}^n$.

If I take $q=\frak{m}$ (the condition on the radical is automatically satisfied) and $n=1$, then this would imply $p^{(m)} \subseteq p^{(m)}\frak{m}$ (for any associated prime $p$ of $R$ since this would satisfy the condition on dimension and requiring that $R/p$ is regular). Am I missing something here?

-
$R$ is a domain; the only associated prime is $(0)$. – Keerthi Madapusi Pera May 18 '11 at 6:04
For readers like myself who haven't seen the notation before, $\mathfrak{p}^{(m)}$ is defined as $R \cap \mathfrak{p}^m R_\mathfrak{p}$ – Hurkyl May 18 '11 at 6:11
To expand on Keerthi's comment: The Auslander–Buchsbaum theorem (en.wikipedia.org/wiki/Auslander-Buchsbaum_theorem) states that every regular local ring is a unique factorization domain. In fact, it is sort of amusing that you can use this result to prove the Auslander-Buchsbaum theorem - since if $p^{(m)} \subseteq p^{(m)}m$, then by Nakayama's lemma $p^{(m)}= 0$, which implies that $p=0$! – auniket May 18 '11 at 6:33

The short answer to what you're missing is that you ignored the dimension criterion. $R/\mathfrak{m}$ is zero-dimensional, so the only prime ideal $\mathfrak{p}$ to which the the theorem would apply is $\mathfrak{p} = (0)$.
Thanks. I did realize that $R/\frak{m}$ is zero dimensional (which is how I concluded that the dimensiona criterion will hold for any associated prime $p$. What I failed to realize was that this would imply $p=0$. – Ajay Patwardhan May 18 '11 at 13:38