# Localization of regular local ring by prime ideal

Let $\quad R := k[X_1,\ldots,X_n]/(I)\quad$ be quotient of polynomial ring by some prime ideal, $\mathfrak{p} \subset \mathfrak{m}$ two ideals of $R$, where $\mathfrak{p}$ - prime ideal and $\mathfrak{m}$ - maximal ideal containing it. What is the easiest proof that if $R_{\mathfrak{m}}$ (localization of $R$ by $\mathfrak{m}$) is regular local ring then $R_{\mathfrak{p}}$ is also regular local ring.

There is well-known result that if $A$ is regular local ring then so is any of its localizations by prime ideals. There are proof of this in many references, but it is very nonelementary. So can someone indicate an easier proofs or give some references for the special case I given above.

Thank you!

• Why is the proof very non-elementary to you? – Youngsu Aug 2 '17 at 12:27
• Well, "non-elementary" is bit vague notion. But all proofs I've seen use some homological methods and other heavy machinery of commutative algebra. So I'm interested if its can be done easier in my specific example. – user468985 Aug 2 '17 at 12:47
• What is the other heavy machinery of commutaive algebra? I don’t think your example is very special: $(R,m)$ is regular iff its completion $(R^m,mR^m)$ is regular. By the Cohen structure theorem, “any” complete local ring is of type $k[[x_1,\dots,x_n]]/(I)$. – Youngsu Aug 2 '17 at 16:24
• Unfortunately, I do not know an alternative proof which does not use homological algebra. – Youngsu Aug 2 '17 at 16:26

I don't know an elementary proof. I know that $R$ regular $\Rightarrow R_P$ regular is a consequence of Auslander-Buchsbaum-Serre's theorem, which involves indeed homological algebra. See the book "Cohen-Macaulay rings", J.Herzog and W.Bruns, a.k.a. the bible of commutative algebra.