**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