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Assuming the Axiom of Choice, local criteria for zero modules and for injective, surjective, and bijective maps

Statement

Assume the Axiom of Choice.

Let $M$ be a left $R$ -module.

$M = 0$ if and only if $M_{p} = 0$ for every prime ideal $p$ , and this is equivalent to $M_{m} = 0$ for every maximal ideal $m$ .
For an $R$ -module homomorphism $f : M \to N$ , the map $f$ is injective, surjective, or bijective if and only if every prime localisation $f_{p}$ has the same property, and this is equivalent to checking every maximal localisation.

Facts & Assumptions

Given: A commutative ring $R$ , left $R$ -modules $M, N$ , and an $R$ -module homomorphism $f : M \to N$ .

[L1]

Localisation identifies kernels and cokernels: $S^{- 1} (\ker f) ≅ \ker (S^{- 1} f)$ and $S^{- 1} (coker f) ≅ coker (S^{- 1} f)$ (Localisation commutes with kernels images and cokernels).

[L2]

Every proper ideal of a nonzero commutative ring is contained in a maximal ideal (In a nonzero commutative ring, every proper ideal is contained in a maximal ideal).

[L3]

Every maximal ideal of a commutative ring is prime (Every maximal ideal of a commutative ring is prime).

[L4]

The annihilator of $m \in M$ is ${Ann}_{R} (m) = {r \in R : r m = 0}$ (Annihilators, torsion elements and the torsion subset of a module).

[L5]

A localised fraction is zero exactly when one denominator kills its numerator (A localised module fraction is zero exactly when one denominator kills its numerator).

[L6]

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

Proof

technique · direct

1.1

If $M = 0$ , then every localisation of $M$ is $0$ .

given

1.2

Suppose $M_{m} = 0$ for every maximal ideal $m$ and $m \in M$ is nonzero. Then ${Ann}_{R} (m)$ is a proper ideal by [L4], so [L2] gives a maximal ideal $m$ containing it. If $m / 1 = 0$ in $M_{m}$ , [L5] gives $t \notin m$ with $t m = 0$ , so $t \in {Ann}_{R} (m) \subseteq m$ , a contradiction. Hence $M_{m} \neq 0$ , contradicting the hypothesis. Therefore $M = 0$ iff all maximal localisations vanish.

L2L4L5choose

2.1

If all prime localisations vanish then all maximal localisations vanish by [L3], so step 1.2 gives $M = 0$ . Conversely, suppose $M_{p} \neq 0$ for some prime ideal $p$ and choose $m / s \neq 0$ in $M_{p}$ . Then no element outside $p$ annihilates $m$ , or else [L5] would give $m / s = 0$ ; hence ${Ann}_{R} (m) \subseteq p$ . By [L2] choose a maximal ideal $m$ containing ${Ann}_{R} (m)$ . The same zero-criterion argument as in step 1.2 gives $m / 1 \neq 0$ in $M_{m}$ , so maximal-local vanishing would fail. Thus prime-local vanishing and maximal-local vanishing are equivalent.

L2L3L4L5step 1.2choose

3.1

The map $f$ is injective iff $\ker f = 0$ . By [L1], this is equivalent to $\ker (f_{p}) = 0$ for every prime $p$ , and then by step 2.1 to $\ker (f_{m}) = 0$ for every maximal $m$ . So $f$ is injective iff all prime localisations, equivalently all maximal localisations, are injective.

L1L6step 2.1

3.2

The map $f$ is surjective iff $coker f = 0$ . By [L1], this is equivalent to $coker (f_{p}) = 0$ for every prime $p$ , and then by step 2.1 to $coker (f_{m}) = 0$ for every maximal $m$ . So $f$ is surjective iff all prime localisations, equivalently all maximal localisations, are surjective.

L1L6step 2.1

4.1

A map is bijective exactly when it is both injective and surjective, so step 3.1 and step 3.2 give the bijective criterion.

step 3.1step 3.2

5.1

Steps 1.2, 2.1, 3.1, 3.2, and 4.1 prove both claims.

step 1.2step 2.1step 3.1step 3.2step 4.1∎

Depends on

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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, Corollary 5.15 and Proposition 5.16 (standard reference, not scraped)