The Stacks project

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

It's possible I'm just not seeing a route to the inequality 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 elements, any prime of which is minimal over has height by (1) of Tag 10.60.12, and now if we apply the dimension formula Tag 10.104.4 to such a prime , we find that . Thus all irreducible components of have dimension , so . Again, though, perhaps this is an overly complicated approach.

Comment #1981 by on

Hi! OK, I think it works by applying Lemma 10.60.13 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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