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The localised Hom map is an isomorphism for finite free sources

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, let $M$ be a finite free $R$ -module, and let $N$ be a left $R$ -module. Then the natural map

θ_{M, N} : S^{- 1} ⁣ {Hom}_{R} (M, N) ⟶ {Hom}_{S^{- 1} R} (S^{- 1} M, S^{- 1} N)

from There is a natural localisation map on Hom is an isomorphism.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , a finite free $R$ -module $M$ with basis $e_{1}, \dots, e_{n}$ , and a left $R$ -module $N$ .

[L1]

The natural map $θ_{M, N}$ satisfies $θ_{M, N} (f / s) (m / u) = f (m) / (s u)$ (There is a natural localisation map on Hom).

[L2]

A basis means that every element of $M$ has a unique finite linear combination in the basis elements (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L3]

A homomorphism from a finite free module is uniquely determined by its basis values, and every tuple of target values occurs as those basis values (For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$ ).

[L4]

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

[L5]

The group ${Hom}_{R} (M, N)$ is an $R$ -module (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L6]

In $S^{- 1} M$ , addition and scalar multiplication satisfy $m / u = \sum_{i} (a_{i} / u) (e_{i} / 1)$ whenever $m = \sum_{i} a_{i} e_{i}$ in $M$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

By [L2] and [L6], every element of $S^{- 1} M$ is an $S^{- 1} R$ -linear combination of $e_{1} / 1, \dots, e_{n} / 1$ , so two $S^{- 1} R$ -linear maps $S^{- 1} M \to S^{- 1} N$ that agree on all $e_{i} / 1$ are equal.

L2L6algebra

1.2

If $θ_{M, N} (f / s) = 0$ , then $f (e_{i}) / s = 0$ for each $i$ , so [L4] gives $u_{i} \in S$ with $u_{i} f (e_{i}) = 0$ . Let $u = u_{1} \dots u_{n}$ . Then $u f (e_{i}) = 0$ for every $i$ , hence $u f = 0$ by [L3]. Since ${Hom}_{R} (M, N)$ is an $R$ -module by [L5], [L4] applied there gives $f / s = 0$ . Thus $θ_{M, N}$ is injective.

L1L3L4L5choose

1.3

Let $φ : S^{- 1} M \to S^{- 1} N$ be $S^{- 1} R$ -linear. Write $φ (e_{i} / 1) = n_{i} / s_{i}$ with $n_{i} \in N$ and $s_{i} \in S$ , set $s = s_{1} \dots s_{n}$ , and set $m_{i} = (s / s_{i}) n_{i}$ . By [L3], there is a unique $R$ -linear map $f : M \to N$ with $f (e_{i}) = m_{i}$ for all $i$ .

L3choose

2.1

For each basis vector, $θ_{M, N} (f / s) (e_{i} / 1) = f (e_{i}) / s = (s / s_{i}) n_{i} / s = n_{i} / s_{i} = φ (e_{i} / 1)$ , so step 1.1 gives $θ_{M, N} (f / s) = φ$ . Thus $θ_{M, N}$ is surjective.

L1step 1.1step 1.3algebra

3.1

Steps 1.2 and 2.1 prove that $θ_{M, N}$ is an isomorphism.

step 1.2step 2.1∎

Depends on

Used by

Localisation of Hom for finite and finitely presented modules Theorem

Dependency tree · two levels

18 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)