Lemma 10.137.5. Let $k$ be a field. Let $S$ be a smooth $k$-algebra. Then $S$ is a local complete intersection.

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

$\xymatrix{ I \ar[r] \ar[d] & I/I^2 \ar[rr] \ar[d] & & I/\mathfrak m'I \ar[d] \\ \Omega _{k[x_1, \ldots , x_ n]/k} \ar[r] & \bigoplus S\text{d}x_ i \ar[rr]^{\text{d}x_ i \mapsto x_ i - a_ i} & & \mathfrak m'/(\mathfrak m')^2 }$

(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 $k$. $\square$

Comment #721 by Keenan Kidwell on

In the third line of the proof, the containment $\mathfrak{m}^\prime\subseteq I$ goes the other way.

Comment #6688 by WhatJiaranEatsTonight on

I think the commutativity has an intuitive explanation. Let f be a polynomial that represents an element in $I/I^2$. The composition of the left and the bottom takes f to $\sum \partial f/\partial x_i(x_i -a_i)$.

Notice that the Taylor expansion of f has vanishing constant since $I\subset m$. And the higher items lie in $m^2$. Thus f is equal to the first order items of its Taylor expansion, which is exactly the form as above.

Comment #6894 by on

@#6688: I sort of agree, but I am not sure it is worth editing what we have now. Can somebody suggest a better way of writing the argument?

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