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 βAβ 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