Lemma 16.11.2. Let $k$ be a field of characteristic $p > 0$. Let $\Lambda$ be a Noetherian geometrically regular $k$-algebra. Let $\mathfrak q \subset \Lambda$ be a prime ideal. Let $n \geq 1$ be an integer and let $E \subset \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$ be a finite subset. Then we can find $m \geq 0$ and $\varphi : k[y_1, \ldots , y_ m] \to \Lambda$ with the following properties

1. setting $\mathfrak p = \varphi ^{-1}(\mathfrak q)$ we have $\mathfrak q\Lambda _\mathfrak q = \mathfrak p \Lambda _\mathfrak q$ and $k[y_1, \ldots , y_ m]_\mathfrak p \to \Lambda _\mathfrak q$ is flat,

2. there is a factorization by homomorphisms of local Artinian rings

$k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to D \to \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$

where the first arrow is essentially smooth and the second is flat,

3. $E$ is contained in $D$ modulo $\mathfrak q^ n\Lambda _\mathfrak q$.

Proof. Set $\bar\Lambda = \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$. Note that $\dim H_1(L_{\kappa (\mathfrak q)/k}) < \infty$ by More on Algebra, Proposition 15.35.1. Pick $A \subset \bar\Lambda$ containing $E$ such that $A$ is local Artinian, essentially of finite type over $k$, the map $A \to \bar\Lambda$ is flat, and $\mathfrak m_ A$ generates the maximal ideal of $\bar\Lambda$, see Lemma 16.11.1. Denote $F = A/\mathfrak m_ A$ the residue field so that $k \subset F \subset K$. Pick $\lambda _1, \ldots , \lambda _ t \in \Lambda$ which map to elements of $A$ in $\bar\Lambda$ such that moreover the images of $\text{d}\lambda _1, \ldots , \text{d}\lambda _ t$ form a basis of $\Omega _{F/k}$. Consider the map $\varphi ' : k[y_1, \ldots , y_ t] \to \Lambda$ sending $y_ j$ to $\lambda _ j$. Set $\mathfrak p' = (\varphi ')^{-1}(\mathfrak q)$. By More on Algebra, Lemma 15.35.2 the ring map $k[y_1, \ldots , y_ t]_{\mathfrak p'} \to \Lambda _\mathfrak q$ is flat and $\Lambda _\mathfrak q/\mathfrak p' \Lambda _\mathfrak q$ is regular. Thus we can choose further elements $\lambda _{t + 1}, \ldots , \lambda _ m \in \Lambda$ which map into $A \subset \bar\Lambda$ and which map to a regular system of parameters of $\Lambda _\mathfrak q/\mathfrak p' \Lambda _\mathfrak q$. We obtain $\varphi : k[y_1, \ldots , y_ m] \to \Lambda$ having property (1) such that $k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to \bar\Lambda$ factors through $A$. Thus $k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to A$ is flat by Algebra, Lemma 10.39.9. By construction the residue field extension $F/\kappa (\mathfrak p)$ is finitely generated and $\Omega _{F/\kappa (\mathfrak p)} = 0$. Hence it is finite separable by More on Algebra, Lemma 15.34.1. Thus $k[y_1, \ldots , y_ m]_\mathfrak p/\mathfrak p^ n k[y_1, \ldots , y_ m]_\mathfrak p \to A$ is finite by Algebra, Lemma 10.54.4. Finally, we conclude that it is étale by Algebra, Lemma 10.143.7. Since an étale ring map is certainly essentially smooth we win. $\square$

Comment #8618 by Nick Addington on

"Essentially smooth" is used here and in the next two lemmas. I guess it means formally smooth and essentially of finite type, or maybe essentially of finite presentation, or (equivalently?) it's a localization of a smooth map; but the definition doesn't seem to be in the stacks project.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).