Lemma 16.10.1 (Ogoma). Let $A$ be a Noetherian ring and let $M$ be a finite $A$-module. Let $S \subset A$ be a multiplicative set. If $\pi \in A$ and $\mathop{\mathrm{Ker}}(\pi : S^{-1}M \to S^{-1}M) = \mathop{\mathrm{Ker}}(\pi ^2 : S^{-1}M \to S^{-1}M)$ then there exists an $s \in S$ such that for any $n > 0$ we have $\mathop{\mathrm{Ker}}(s^ n\pi : M \to M) = \mathop{\mathrm{Ker}}((s^ n\pi )^2 : M \to M)$.

## 16.10 Separable residue fields

In this section we explain how to solve a local problem in the case of a separable residue field extension.

**Proof.**
Let $K = \mathop{\mathrm{Ker}}(\pi : M \to M)$ and $K' = \{ m \in M \mid \pi ^2 m = 0\text{ in }S^{-1}M\} $ and $Q = K'/K$. Note that $S^{-1}Q = 0$ by assumption. Since $A$ is Noetherian we see that $Q$ is a finite $A$-module. Hence we can find an $s \in S$ such that $s$ annihilates $Q$. Then $s$ works.
$\square$

Lemma 16.10.2. Let $\Lambda $ be a Noetherian ring. Let $I \subset \Lambda $ be an ideal. Let $I \subset \mathfrak q$ be a prime. Let $n, e$ be positive integers Assume that $\mathfrak q^ n\Lambda _\mathfrak q \subset I\Lambda _\mathfrak q$ and that $\Lambda _\mathfrak q$ is a regular local ring of dimension $d$. Then there exists an $n > 0$ and $\pi _1, \ldots , \pi _ d \in \Lambda $ such that

$(\pi _1, \ldots , \pi _ d)\Lambda _\mathfrak q = \mathfrak q\Lambda _\mathfrak q$,

$\pi _1^ n, \ldots , \pi _ d^ n \in I$, and

for $i = 1, \ldots , d$ we have

\[ \text{Ann}_{\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)\Lambda }(\pi _ i) = \text{Ann}_{\Lambda /(\pi _1^ e, \ldots , \pi _{i - 1}^ e)\Lambda }(\pi _ i^2). \]

**Proof.**
Set $S = \Lambda \setminus \mathfrak q$ so that $\Lambda _\mathfrak q = S^{-1}\Lambda $. First pick $\pi _1, \ldots , \pi _ d$ with (1) which is possible as $\Lambda _\mathfrak q$ is regular. By assumption $\pi _ i^ n \in I\Lambda _\mathfrak q$. Thus we can find $s_1, \ldots , s_ d \in S$ such that $s_ i\pi _ i^ n \in I$. Replacing $\pi _ i$ by $s_ i\pi _ i$ we get (2). Note that (1) and (2) are preserved by further multiplying by elements of $S$. Suppose that (3) holds for $i = 1, \ldots , t$ for some $t \in \{ 0, \ldots , d\} $. Note that $\pi _1, \ldots , \pi _ d$ is a regular sequence in $S^{-1}\Lambda $, see Algebra, Lemma 10.105.3. In particular $\pi _1^ e, \ldots , \pi _ t^ e, \pi _{t + 1}$ is a regular sequence in $S^{-1}\Lambda = \Lambda _\mathfrak q$ by Algebra, Lemma 10.67.9. Hence we see that

Thus we get (3) for $i = t + 1$ after replacing $\pi _{t + 1}$ by $s\pi _{t + 1}$ for some $s \in S$ by Lemma 16.10.1. By induction on $t$ this produces a sequence satisfying (1), (2), and (3). $\square$

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

$k$ is a field,

$\Lambda $ is Noetherian,

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

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

the field extension $k \subset \kappa (\mathfrak q)$ 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

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

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

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 10.127.2. Hence

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 10.152.11 and Proposition 16.5.3. Existence of the desired solution follows from Algebra, Lemma 10.126.4. $\square$

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