Lemma 10.135.3 (00SB)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.135: Local complete intersections
Lemma 10.135.3 (cite)

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Lemma 10.135.3. Let $k$ be a field. Let $S$ be a finite type $k$ -algebra. If $S$ is a local complete intersection, then $S$ is a Cohen-Macaulay ring.

Proof. Choose a maximal prime $\mathfrak m$ of $S$ . We have to show that $S_\mathfrak m$ is Cohen-Macaulay. By assumption we may assume $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $\dim (S) = n - c$ . Let $\mathfrak m' \subset k[x_1, \ldots , x_ n]$ be the maximal ideal corresponding to $\mathfrak m$ . According to Proposition 10.114.2 the local ring $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ is regular local of dimension $n$ . In particular it is Cohen-Macaulay by Lemma 10.106.3. By Lemma 10.60.13 applied $c$ times the local ring $S_{\mathfrak m} = k[x_1, \ldots , x_ n]_{\mathfrak m'}/(f_1, \ldots , f_ c)$ has dimension $\geq n - c$ . By assumption $\dim (S_{\mathfrak m}) \leq n - c$ . Thus we get equality. This implies that $f_1, \ldots , f_ c$ is a regular sequence in $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ and that $S_{\mathfrak m}$ is Cohen-Macaulay, see Proposition 10.103.4. $\square$

Comments (2)

Comment #1908 by Keenan Kidwell on April 02, 2016 at 23:49

It's possible I'm just not seeing a route to the inequality $\dim(S_\mathfrak{m})\geq n-c$ which is contained entirely in Tag 10.60, but the only argument I can think of for this uses more than what's in the section on dimension and perhaps a couple precise references would be good. Namely, since we're taking the quotient of a Noetherian ring by an ideal generated by $c$ elements, any prime of $k[x_1,\ldots,x_n]_{\mathfrak{m}^\prime}$ which is minimal over $(f_1,\ldots,f_c)$ has height $\leq c$ by (1) of Tag 10.60.12, and now if we apply the dimension formula Tag 10.104.4 to such a prime $\mathfrak{p}$ , we find that $\dim(k[x_1,\ldots,x_n]_{\mathfrak{m}^\prime}/\mathfrak{p})\geq n-c$ . Thus all irreducible components of $\mathrm{Spec(S_\mathfrak{m})}$ have dimension $\geq n-c$ , so $\dim(S_\mathfrak{m})\geq n-c$ . Again, though, perhaps this is an overly complicated approach.

Comment #1981 by Johan on May 03, 2016 at 12:32

Hi! OK, I think it works by applying Lemma 10.60.13 $c$ times. The corresponding edit is here. I guess maybe we should restate Lemma 10.60.13 in terms of sequences of elements.

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