The Stacks project

Lemma 10.140.2. Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. The following are equivalent:

  1. The ring $S_{\mathfrak m}$ is a regular local ring.

  2. We have $\dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) \leq \dim (S_{\mathfrak m})$.

  3. We have $\dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) = \dim (S_{\mathfrak m})$.

  4. There exists a $g \in S$, $g \not\in \mathfrak m$ such that $S_ g$ is smooth over $k$. In other words $S/k$ is smooth at $\mathfrak m$.

Proof. Note that (1), (2) and (3) are equivalent by Lemma 10.140.1 and Definition 10.110.7.

Assume that $S$ is smooth at $\mathfrak m$. By Lemma 10.137.10 we see that $S_ g$ is standard smooth over $k$ for a suitable $g \in S$, $g \not\in \mathfrak m$. Hence by Lemma 10.137.7 we see that $\Omega _{S_ g/k}$ is free of rank $\dim (S_ g)$. Hence by Lemma 10.140.1 we see that $\dim (S_{\mathfrak m}) = \dim (\mathfrak m/\mathfrak m^2)$ in other words $S_\mathfrak m$ is regular.

Conversely, suppose that $S_{\mathfrak m}$ is regular. Let $d = \dim (S_{\mathfrak m}) = \dim \mathfrak m/\mathfrak m^2$. Choose a presentation $S = k[x_1, \ldots , x_ n]/I$ such that $x_ i$ maps to an element of $\mathfrak m$ for all $i$. In other words, $\mathfrak m'' = (x_1, \ldots , x_ n)$ is the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. Note that we have a short exact sequence

\[ I/\mathfrak m''I \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m/(\mathfrak m)^2 \to 0 \]

Pick $c = n - d$ elements $f_1, \ldots , f_ c \in I$ such that their images in $\mathfrak m''/(\mathfrak m'')^2$ span the kernel of the map to $\mathfrak m/\mathfrak m^2$. This is clearly possible. Denote $J = (f_1, \ldots , f_ c)$. So $J \subset I$. Denote $S' = k[x_1, \ldots , x_ n]/J$ so there is a surjection $S' \to S$. Denote $\mathfrak m' = \mathfrak m''S'$ the corresponding maximal ideal of $S'$. Hence we have

\[ \xymatrix{ k[x_1, \ldots , x_ n] \ar[r] & S' \ar[r] & S \\ \mathfrak m'' \ar[u] \ar[r] & \mathfrak m' \ar[r] \ar[u] & \mathfrak m \ar[u] } \]

By our choice of $J$ the exact sequence

\[ J/\mathfrak m''J \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m'/(\mathfrak m')^2 \to 0 \]

shows that $\dim ( \mathfrak m'/(\mathfrak m')^2 ) = d$. Since $S'_{\mathfrak m'}$ surjects onto $S_{\mathfrak m}$ we see that $\dim (S_{\mathfrak m'}) \geq d$. Hence by the discussion preceding Definition 10.60.10 we conclude that $S'_{\mathfrak m'}$ is regular of dimension $d$ as well. Because $S'$ was cut out by $c = n - d$ equations we conclude that there exists a $g' \in S'$, $g' \not\in \mathfrak m'$ such that $S'_{g'}$ is a global complete intersection over $k$, see Lemma 10.135.4. Also the map $S'_{\mathfrak m'} \to S_{\mathfrak m}$ is a surjection of Noetherian local domains of the same dimension and hence an isomorphism. Hence $S' \to S$ is surjective with finitely generated kernel and becomes an isomorphism after localizing at $\mathfrak m'$. Thus we can find $g' \in S'$, $g \not\in \mathfrak m'$ such that $S'_{g'} \to S_{g'}$ is an isomorphism. All in all we conclude that after replacing $S$ by a principal localization we may assume that $S$ is a global complete intersection.

At this point we may write $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $\dim S = n - c$. Recall that the naive cotangent complex of this algebra is given by

\[ \bigoplus S \cdot f_ j \to \bigoplus S \cdot \text{d}x_ i \]

see Lemma 10.136.12. By Lemma 10.137.16 in order to show that $S$ is smooth at $\mathfrak m$ we have to show that one of the $c \times c$ minors $g_ I$ of the matrix “$A$” giving the map above does not vanish at $\mathfrak m$. By Lemma 10.140.1 the matrix $A \bmod \mathfrak m$ has rank $c$. Thus we win. $\square$

Comments (1)

Comment #725 by Keenan Kidwell on

For the description of the naive cotangent complex in the last paragraph, I think the correct tag is (3) of 00SV. This reference is cited for the same reason in the proof of 00TE, but that's not where it is proved. For "smooth at iff some minor is non-zero modulo ," 00TE is the right tag.

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