The Stacks project

Lemma 16.11.3. Let $\varphi : k[y_1, \ldots , y_ m] \to \Lambda $, $n$, $\mathfrak q$, $\mathfrak p$ and

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

be as in Lemma 16.11.2. Then for any $\lambda \in \Lambda \setminus \mathfrak q$ there exists an integer $q > 0$ and a factorization

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

such that $D \to D'$ is an essentially smooth map of local Artinian rings, the last arrow is flat, and $\lambda ^ q$ is in $D'$.

Proof. Set $\bar\Lambda = \Lambda _\mathfrak q/\mathfrak q^ n\Lambda _\mathfrak q$. Let $\bar\lambda $ be the image of $\lambda $ in $\bar\Lambda $. Let $\alpha \in \kappa (\mathfrak q)$ be the image of $\lambda $ in the residue field. Let $k \subset F \subset \kappa (\mathfrak q)$ be the residue field of $D$. If $\alpha $ is in $F$ then we can find an $x \in D$ such that $x \bar\lambda = 1 \bmod \mathfrak q$. Hence $(x \bar\lambda )^ q = 1 \bmod (\mathfrak q)^ q$ if $q$ is divisible by $p$. Hence $\bar\lambda ^ q$ is in $D$. If $\alpha $ is transcendental over $F$, then we can take $D' = (D[\bar\lambda ])_\mathfrak m$ equal to the subring generated by $D$ and $\bar\lambda $ localized at $\mathfrak m = D[\bar\lambda ] \cap \mathfrak q \bar\Lambda $. This works because $D[\bar\lambda ]$ is in fact a polynomial algebra over $D$ in this case. Finally, if $\lambda \bmod \mathfrak q$ is algebraic over $F$, then we can find a $p$-power $q$ such that $\alpha ^ q$ is separable algebraic over $F$, see Fields, Section 9.28. Note that $D$ and $\bar\Lambda $ are henselian local rings, see Algebra, Lemma 10.153.10. Let $D \to D'$ be a finite ├ętale extension whose residue field extension is $F(\alpha ^ q)/F$, see Algebra, Lemma 10.153.7. Since $\bar\Lambda $ is henselian and $F(\alpha ^ q)$ is contained in its residue field we can find a factorization $D' \to \bar\Lambda $. By the first part of the argument we see that $\bar\lambda ^{qq'} \in D'$ for some $q' > 0$. $\square$

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