TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Assuming the Axiom of Choice, a sequence of modules is exact exactly when all prime localisations are exact

Statement

Assume the Axiom of Choice.

Let

M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''}

be a sequence of $R$ -module homomorphisms with $g \circ f = 0$ . Then the sequence is exact at $M$ if and only if, for every prime ideal $p$ , the localised sequence

M_{p}^{'} \overset{f_{p}}{\to} M_{p} \overset{g_{p}}{\to} M_{p}^{''}

is exact at $M_{p}$ . Equivalently, it suffices to check exactness at every maximal ideal.

Facts & Assumptions

Given: A commutative ring $R$ and a sequence $M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''}$ of left $R$ -modules with $g \circ f = 0$ .

[L1]

Exactness at $M$ means $im f = \ker g$ (Exact sequences and short exact sequences of modules).

[L2]

Localisation identifies kernels and images, and it commutes with quotient modules (Localisation commutes with kernels images and cokernels, Localisation commutes with quotient modules and arbitrary direct sums).

[L3]

A module is zero exactly when all of its prime localisations are zero, equivalently all of its maximal localisations are zero (Assuming the Axiom of Choice, local criteria for zero modules and for injective, surjective, and bijective maps).

[L4]

The image and kernel are the standard submodule constructions attached to a module homomorphism (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Because $g \circ f = 0$ , one has $im f \subseteq \ker g$ , so the quotient module $H : = \ker g / im f$ is defined. By [L1], the original sequence is exact at $M$ if and only if $H = 0$ .

L1L4

2.1

For every prime ideal $p$ , [L2] gives $H_{p} ≅ (\ker g)_{p} / (im f)_{p} ≅ \ker (g_{p}) / im (f_{p})$ . Therefore the localised sequence is exact at $M_{p}$ if and only if $H_{p} = 0$ .

L2step 1.1

3.1

By [L3], $H = 0$ if and only if $H_{p} = 0$ for every prime ideal $p$ , and this is equivalent to $H_{m} = 0$ for every maximal ideal $m$ . Combining this with steps 1.1 and 2.1 gives the prime-local and maximal-local exactness criteria.

L3step 1.1step 2.1

4.1

Step 3.1 proves the theorem.

step 3.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Proposition 13.43 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 5.16 (standard reference, not scraped)