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Inverse limits preserve kernels

Statement

Let

(f_{n}) ⁣ : (M_{n}, φ_{n}) \to (N_{n}, ψ_{n})

be a morphism of inverse systems of $R$ -modules. Then the kernel of the induced map

\underset{\leftarrow}{l i m} M_{n} ⟶ \underset{\leftarrow}{l i m} N_{n}

is canonically isomorphic to

\underset{\leftarrow}{l i m} \ker (f_{n}) .

Consequently, if

0 \to A_{n} \to B_{n} \to C_{n}

is an exact sequence of inverse systems, 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}

is exact.

Facts & Assumptions

Given: A morphism of inverse systems $(f_{n}) ⁣ : (M_{n}, φ_{n}) \to (N_{n}, ψ_{n})$ .

[L1]

The inverse limit consists of the compatible elements in the product system (Inverse systems and inverse limits of modules).

[L2]

The kernel of a homomorphism is the submodule of elements mapping to $0$ (Module homomorphism and isomorphism, kernel, image and cokernel).

[L3]

A map into an inverse limit is determined by its compatible coordinate maps (Universal property of an inverse limit of modules).

Proof

technique · direct

1.1

For each $n$ , let $K_{n} : = \ker (f_{n})$ . Compatibility of the squares $ψ_{n} \circ f_{n} = f_{n - 1} \circ φ_{n}$ shows that $φ_{n} (K_{n}) \subseteq K_{n - 1}$ , so the kernels form an inverse system.

L2given

1.2

An element $(x_{n}) \in \underset{\leftarrow}{l i m} M_{n}$ lies in the kernel of $\underset{\leftarrow}{l i m} M_{n} \to \underset{\leftarrow}{l i m} N_{n}$ exactly when $f_{n} (x_{n}) = 0$ for every $n$ . By [L2], that is equivalent to $x_{n} \in K_{n}$ for every $n$ . Together with the compatibility condition from [L1], this says precisely that $(x_{n}) \in \underset{\leftarrow}{l i m} K_{n}$ .

L1L2

2.1

Thus the underlying subsets of $\ker ⁣ (\underset{\leftarrow}{l i m} M_{n} \to \underset{\leftarrow}{l i m} N_{n}) and \underset{\leftarrow}{l i m} \ker (f_{n})$ coincide inside $\prod M_{n}$ , and the module structures also agree componentwise. Hence these two modules are canonically equal, in particular canonically isomorphic.

step 1.2

3.1

For an exact sequence $0 \to A_{n} \to B_{n} \to C_{n}$ , exactness means $A_{n} = \ker (B_{n} \to C_{n})$ for every $n$ . Applying step 2.1 to the maps $B_{n} \to C_{n}$ gives $\ker ⁣ (\underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n}) = \underset{\leftarrow}{l i m} A_{n},$ which is exactly left exactness.

step 2.1L3∎

Depends on

Used by

Countable Mittag-Leffler systems preserve short exactness on inverse limits Theorem

Dependency tree · two levels

6 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

J. S. Milne, A Primer of Commutative Algebra, Proposition 9.6 (standard reference, not scraped)
The Stacks Project, Section 10.87 (standard reference, not scraped)