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May 18, 2011 at 13:39 vote accept Ajay Patwardhan
May 18, 2011 at 12:14 answer added Sean Sather-Wagstaff timeline score: 4
May 18, 2011 at 7:02 answer added user13113 timeline score: 3
May 18, 2011 at 6:33 comment added pinaki 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 $p=0$!
May 18, 2011 at 6:11 comment added user13113 For readers like myself who haven't seen the notation before, $\mathfrak{p}^{(m)}$ is defined as $R \cap \mathfrak{p}^m R_\mathfrak{p}$
May 18, 2011 at 6:04 comment added Keerthi Madapusi $R$ is a domain; the only associated prime is $(0)$.
May 18, 2011 at 4:51 history asked Ajay Patwardhan CC BY-SA 3.0