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A finite presentation reduces localised Hom to the finite free case

Statement

Let $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M ⟶ 0$ be a finite presentation of an $R$ -module $M$ , and let $N$ be an $R$ -module. If the natural localisation maps for $R^{n}$ and $R^{m}$ are isomorphisms, then the natural localisation map $S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ is an isomorphism as well.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , an $R$ -module $N$ , and a finite presentation $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ .

[L1]

A finite presentation is an exact sequence $R^{m} \overset{α}{\to} R^{n} \overset{β}{\to} M \to 0$ , and equivalently $M ≅ R^{n} / im α$ (Finitely presented modules and finitely presented algebras).

[L2]

Localisation sends short exact sequences to short exact sequences (Localisation of modules is exact).

[L3]

The localisation map on Hom is natural in the source, with formula $θ_{X, N} (f / s) (x / u) = f (x) / (s u)$ (There is a natural localisation map on Hom).

Proof

technique · direct

1.1L1algebra

Because $β$ is surjective in [L1], a homomorphism $h : R^{n} \to N$ factors through $M$ exactly when $h$ kills $im α$ , equivalently when $h \circ α = 0$ . Thus ${Hom}_{R} (M, N)$ is the kernel of $α^{*} : {Hom}_{R} (R^{n}, N) \to {Hom}_{R} (R^{m}, N)$ , $h \mapsto h \circ α$ .

1.2L1L2algebra

By [L2], localising the presentation of [L1] gives an exact sequence $S^{- 1} R^{m} \overset{S^{- 1} α}{\to} S^{- 1} R^{n} \overset{S^{- 1} β}{\to} S^{- 1} M \to 0$ , so the same argument identifies ${Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ with the kernel of $(S^{- 1} α)^{*}$ .

1.3L3algebra

Naturality from [L3] makes the square between $α^{*}$ and $(S^{- 1} α)^{*}$ commute.

2.1step 1.1step 1.2step 1.3algebra

If the two vertical maps in step 1.3 are isomorphisms, then they identify the kernel in step 1.1 with the kernel in step 1.2. Therefore the induced map on those kernels, namely $S^{- 1} ⁣ {Hom}_{R} (M, N) \to {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)$ , is an isomorphism.

3.1step 2.1∎

Step 2.1 is exactly the reduction from a finite presentation to the finite free case.

Depends on

Used by

Localisation of Hom for finite and finitely presented modules Theorem

Dependency tree · two levels

16 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., Proposition 12.25 (standard reference, not scraped)