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Countable Mittag-Leffler systems preserve short exactness on inverse limits

Statement

Assume the Axiom of Choice.

Let

0 \to A_{n} \overset{f_{n}}{\to} B_{n} \overset{g_{n}}{\to} C_{n} \to 0

be a short exact sequence of inverse systems of $R$ -modules indexed by $N_{\geq 1}$ . If $(A_{n})$ is Mittag-Leffler, then

0 \to \underset{\leftarrow}{l i m} A_{n} \to \underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n} \to 0

is exact.

Facts & Assumptions

Given: A short exact sequence of inverse systems $0 \to A_{n} \overset{f_{n}}{\to} B_{n} \overset{g_{n}}{\to} C_{n} \to 0$ with $(A_{n})$ Mittag-Leffler.

[L1]

Inverse limits are left exact (Inverse limits preserve kernels).

[L2]

The Mittag-Leffler condition means that for each fixed stage $m$ , the images of the transition maps into $A_{m}$ eventually stabilize (Mittag-Leffler inverse systems).

Proof

technique · direct

1.1

By [L1], the sequence of inverse limits is already exact at $\underset{\leftarrow}{l i m} A_{n}$ and at $\underset{\leftarrow}{l i m} B_{n}$ . It remains to prove surjectivity of $\underset{\leftarrow}{l i m} B_{n} ⟶ \underset{\leftarrow}{l i m} C_{n} .$

1.2

Let $c = (c_{n})_{n \geq 1} \in \underset{\leftarrow}{l i m} C_{n}$ . For each $n$ , set $E_{n} : = g_{n}^{- 1} (c_{n}) \subseteq B_{n} .$ Since $g_{n}$ is surjective, each $E_{n}$ is nonempty. Compatibility of the inverse system and of the family $(c_{n})$ makes every transition map $B_{n + 1} \to B_{n}$ restrict to a map $E_{n + 1} \to E_{n}$ .

givenconstruct

1.3

The system $(E_{n})$ is Mittag-Leffler as a system of sets. Fix $m$ . By [L2] choose $c (m) \geq m$ such that $im (A_{n} \to A_{m}) = im (A_{c (m)} \to A_{m}) (n \geq c (m)) .$ For $n \geq c (m)$ the inclusion $im (E_{n} \to E_{m}) \subseteq im (E_{c (m)} \to E_{m})$ is automatic. For the reverse inclusion, take $y \in im (E_{c (m)} \to E_{m})$ and choose $e_{c} \in E_{c (m)}$ mapping to $y$ . Choose any $e_{n}^{0} \in E_{n}$ . The images of $e_{c}$ and $e_{n}^{0}$ in $C_{c (m)}$ are both $c_{c (m)}$ , so their difference lies in $A_{c (m)}$ . By the stabilization choice there is $a_{n} \in A_{n}$ whose image in $A_{m}$ equals the image of $e_{c} - e_{n}^{0}$ . Then $e_{n} : = e_{n}^{0} + a_{n}$ lies in $E_{n}$ and maps to $y$ in $E_{m}$ . Hence the images stabilize.

L2choosealgebra

2.1

For each $n$ , let $E_{n}^{'} : = ⋂_{m \geq n} im (E_{m} \to E_{n}) .$ Because $(E_{n})$ is Mittag-Leffler and nonempty, $E_{n}^{'}$ is equal to one stable image and is therefore nonempty. The restricted maps $E_{n + 1}^{'} \to E_{n}^{'}$ are surjective: if $y \in E_{n}^{'}$ , then $y$ comes from some sufficiently high stage $E_{m}$ with $m \geq n + 1$ , and the image of that same element in $E_{n + 1}$ lies in $E_{n + 1}^{'}$ and maps to $y$ .

step 1.3construct

3.1

By the Axiom of Choice, choose $x_{1} \in E_{1}^{'}$ , and after $x_{n}$ has been chosen choose $x_{n + 1} \in E_{n + 1}^{'}$ mapping to $x_{n}$ ; this is possible by surjectivity from step 2.1. Then $(x_{n})_{n \geq 1}$ is an element of $\underset{\leftarrow}{l i m} E_{n}$ , hence of $\underset{\leftarrow}{l i m} B_{n}$ , and by construction it maps to $c \in \underset{\leftarrow}{l i m} C_{n}$ .

step 2.1choose

4.1

Therefore $\underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n}$ is surjective. Combined with step 1.1, this proves exactness of $0 \to \underset{\leftarrow}{l i m} A_{n} \to \underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n} \to 0.$

step 1.1step 3.1∎

Depends on

Used by

Adic completion is exact on finite modules over a Noetherian ring Theorem

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Lemma 22.7 (standard reference, not scraped)
The Stacks Project, Lemma 10.86.4 (standard reference, not scraped)