Lemma 16.10.3. Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation 16.9.1 where

1. $k$ is a field,

2. $\Lambda$ is Noetherian,

3. $\mathfrak q$ is minimal over $\mathfrak h_ A$,

4. $\Lambda _\mathfrak q$ is a regular local ring, and

5. the field extension $\kappa (\mathfrak q)/k$ is separable.

Then $k \to A \to \Lambda \supset \mathfrak q$ can be resolved.

Proof. Set $d = \dim \Lambda _\mathfrak q$. Set $R = k[x_1, \ldots , x_ d]$. Choose $n > 0$ such that $\mathfrak q^ n\Lambda _\mathfrak q \subset \mathfrak h_ A\Lambda _\mathfrak q$ which is possible as $\mathfrak q$ is minimal over $\mathfrak h_ A$. Choose generators $a_1, \ldots , a_ r$ of $H_{A/R}$. Set

$B = A[x_1, \ldots , x_ d, z_{ij}]/(x_ i^ n - \sum z_{ij}a_ j)$

Each $B_{a_ j}$ is smooth over $R$ it is a polynomial algebra over $A_{a_ j}[x_1, \ldots , x_ d]$ and $A_{a_ j}$ is smooth over $k$. Hence $B_{x_ i}$ is smooth over $R$. Let $B \to C$ be the $R$-algebra map constructed in Lemma 16.3.1 which comes with a $R$-algebra retraction $C \to B$. In particular a map $C \to \Lambda$ fitting into the diagram above. By construction $C_{x_ i}$ is a smooth $R$-algebra with $\Omega _{C_{x_ i}/R}$ free. Hence we can find $c > 0$ such that $x_ i^ c$ is strictly standard in $C/R$, see Lemma 16.3.7. Now choose $\pi _1, \ldots , \pi _ d \in \Lambda$ as in Lemma 16.10.2 where $n = n$, $e = 8c$, $\mathfrak q = \mathfrak q$ and $I = \mathfrak h_ A$. Write $\pi _ i^ n = \sum \lambda _{ij} a_ j$ for some $\pi _{ij} \in \Lambda$. There is a map $B \to \Lambda$ given by $x_ i \mapsto \pi _ i$ and $z_{ij} \mapsto \lambda _{ij}$. Set $R = k[x_1, \ldots , x_ d]$. Diagram

$\xymatrix{ R \ar[r] & B \ar[rd] \\ k \ar[u] \ar[r] & A \ar[u] \ar[r] & \Lambda }$

Now we apply Lemma 16.9.2 to $R \to C \to \Lambda \supset \mathfrak q$ and the sequence of elements $x_1^ c, \ldots , x_ d^ c$ of $R$. Assumption (2) is clear. Assumption (1) holds for $R$ by inspection and for $\Lambda$ by our choice of $\pi _1, \ldots , \pi _ d$. (Note that if $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^2)$, then we have $\text{Ann}_\Lambda (\pi ) = \text{Ann}_\Lambda (\pi ^ c)$ for all $c > 0$.) Thus it suffices to resolve

$R/(x_1^ e, \ldots , x_ d^ e) \to C/(x_1^ e, \ldots , x_ d^ e) \to \Lambda /(\pi _1^ e, \ldots , \pi _ d^ e) \supset \mathfrak q/(\pi _1^ e, \ldots , \pi _ d^ e)$

for $e = 8c$. By Lemma 16.9.4 it suffices to resolve this after localizing at $\mathfrak q$. But since $x_1, \ldots , x_ d$ map to a regular sequence in $\Lambda _\mathfrak q$ we see that $R_\mathfrak p \to \Lambda _\mathfrak q$ is flat, see Algebra, Lemma 10.128.2. Hence

$R_\mathfrak p/(x_1^ e, \ldots , x_ d^ e) \to \Lambda _\mathfrak q/(\pi _1^ e, \ldots , \pi _ d^ e)$

is a flat ring map of Artinian local rings. Moreover, this map induces a separable field extension on residue fields by assumption. Thus this map is a filtered colimit of smooth algebras by Algebra, Lemma 10.158.11 and Proposition 16.5.3. Existence of the desired solution follows from Algebra, Lemma 10.127.4. $\square$

Comment #2760 by Anonymous on

Towards the end of the proof it should say $R_\mathfrak{p} \rightarrow \Lambda_\mathfrak{q}$ is flat by Algebra, Lemma 10.127.2. And the following map should map from $R_\mathfrak{p}/(x_1^e,\dots,x_d^e)$.

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