Lemma 10.140.3. Let $k$ be any field. Let $S$ be a finite type $k$-algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$. The following are equivalent:

The $k$-algebra $S$ is smooth at $\mathfrak q$ over $k$.

We have $\dim _{\kappa (\mathfrak q)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak q) \leq \dim _ x X$.

We have $\dim _{\kappa (\mathfrak q)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak q) = \dim _ x X$.

Moreover, in this case the local ring $S_{\mathfrak q}$ is regular.

**Proof.**
If $S$ is smooth at $\mathfrak q$ over $k$, then there exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is standard smooth over $k$, see Lemma 10.137.10. A standard smooth algebra over $k$ has a module of differentials which is free of rank equal to the dimension, see Lemma 10.137.7 (use that a relative global complete intersection over a field has dimension equal to the number of variables minus the number of equations). Thus we see that (1) implies (3). To finish the proof of the lemma it suffices to show that (2) implies (1) and that it implies that $S_{\mathfrak q}$ is regular.

Assume (2). By Nakayama's Lemma 10.20.1 we see that $\Omega _{S/k, \mathfrak q}$ can be generated by $\leq \dim _ x X$ elements. We may replace $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ such that $\Omega _{S/k}$ is generated by at most $\dim _ x X$ elements. Let $K/k$ be an algebraically closed field extension such that there exists a $k$-algebra map $\psi : \kappa (\mathfrak q) \to K$. Consider $S_ K = K \otimes _ k S$. Let $\mathfrak m \subset S_ K$ be the maximal ideal corresponding to the surjection

\[ \xymatrix{ S_ K = K \otimes _ k S \ar[r] & K \otimes _ k \kappa (\mathfrak q) \ar[r]^-{\text{id}_ K \otimes \psi } & K. } \]

Note that $\mathfrak m \cap S = \mathfrak q$, in other words $\mathfrak m$ lies over $\mathfrak q$. By Lemma 10.116.6 the dimension of $X_ K = \mathop{\mathrm{Spec}}(S_ K)$ at the point corresponding to $\mathfrak m$ is $\dim _ x X$. By Lemma 10.114.6 this is equal to $\dim ((S_ K)_{\mathfrak m})$. By Lemma 10.131.12 the module of differentials of $S_ K$ over $K$ is the base change of $\Omega _{S/k}$, hence also generated by at most $\dim _ x X = \dim ((S_ K)_{\mathfrak m})$ elements. By Lemma 10.140.2 we see that $S_ K$ is smooth at $\mathfrak m$ over $K$. By Lemma 10.137.18 this implies that $S$ is smooth at $\mathfrak q$ over $k$. This proves (1). Moreover, we know by Lemma 10.140.2 that the local ring $(S_ K)_{\mathfrak m}$ is regular. Since $S_{\mathfrak q} \to (S_ K)_{\mathfrak m}$ is flat we conclude from Lemma 10.110.9 that $S_{\mathfrak q}$ is regular.
$\square$

## Comments (4)

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