The Stacks project

Proof. By Lemmas 10.137.4 and 10.135.11 it suffices to prove this when $k$ is algebraically closed. Choose a presentation $\alpha : k[x_1, \ldots , x_ n] \to S$ with kernel $I$. Let $\mathfrak m$ be a maximal ideal of $S$, and let $\mathfrak m' \supset I$ be the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. We will show that condition (5) of Lemma 10.135.4 holds (with $\mathfrak m$ instead of $\mathfrak q$). We may write $\mathfrak m' = (x_1 - a_1, \ldots , x_ n - a_ n)$ for some $a_ i \in k$, because $k$ is algebraically closed, see Theorem 10.34.1. By our assumption that $k \to S$ is smooth the $S$-module map $\text{d} : I/I^2 \to \bigoplus _{i = 1}^ n S \text{d}x_ i$ is a split injection. Hence the corresponding map $I/\mathfrak m' I \to \bigoplus \kappa (\mathfrak m') \text{d}x_ i$ is injective. Say $\dim _{\kappa (\mathfrak m')}(I/\mathfrak m' I) = c$ and pick $f_1, \ldots , f_ c \in I$ which map to a $\kappa (\mathfrak m')$-basis of $I/\mathfrak m' I$. By Nakayama's Lemma 10.20.1 we see that $f_1, \ldots , f_ c$ generate $I_{\mathfrak m'}$ over $k[x_1, \ldots , x_ n]_{\mathfrak m'}$. Consider the commutative diagram

(proof commutativity omitted). The middle vertical map is the one defining the naive cotangent complex of $\alpha $. Note that the right lower horizontal arrow induces an isomorphism $\bigoplus \kappa (\mathfrak m') \text{d}x_ i \to \mathfrak m'/(\mathfrak m')^2$. Hence our generators $f_1, \ldots , f_ c$ of $I_{\mathfrak m'}$ map to a collection of elements in $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ whose classes in $\mathfrak m'/(\mathfrak m')^2$ are linearly independent over $\kappa (\mathfrak m')$. Therefore they form a regular sequence in the ring $k[x_1, \ldots , x_ n]_{\mathfrak m'}$ by Lemma 10.106.3. This verifies condition (5) of Lemma 10.135.4 hence $S_ g$ is a global complete intersection over $k$ for some $g \in S$, $g \not\in \mathfrak m$. As this works for any maximal ideal of $S$ we conclude that $S$ is a local complete intersection over $k$. $\square$