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Over a Noetherian ring the homomorphism module between two finitely generated modules is finitely generated

Statement

Let $R$ be a Noetherian commutative ring and let $M, N$ be finitely generated $R$ -modules. Then ${Hom}_{R} (M, N)$ , with the $R$ -module structure of Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module, is a finitely generated $R$ -module.

The proof exhibits ${Hom}_{R} (M, N)$ as isomorphic to a submodule of $N^{n}$ for a suitable $n \in N$ . It does not assert that the embedding is onto, and in general it is not.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and finitely generated $R$ -modules $M$ and $N$ .

[L1]

A module is finitely generated when it equals $⟨ S ⟩_{R}$ for some finite subset $S$ (Generated submodule, cyclic and finitely generated modules, module basis and free module).

[L2]

Every set map $u ⁣ : X \to M$ extends uniquely to an $R$ -module homomorphism $\overset{ˉ}{u} ⁣ : R^{(X)} \to M$ with $\overset{ˉ}{u} (e_{x}) = u (x)$ (Universal property of the free module on a set).

[L3]

For a ring $R$ , a left $R$ -module $M$ and $S \subseteq M$ , the submodule $⟨ S ⟩_{R}$ is the set of finite sums $\sum_{i = 1}^{k} r_{i} s_{i}$ with $k \in N$ , $r_{i} \in R$ and $s_{i} \in S$ (The submodule generated by a subset consists of the finite $R$ -linear combinations of that subset).

[L4]

For a homomorphism $u ⁣ : M \to N$ and a module $X$ , precomposition gives a map $u^{*} ⁣ : {Hom}_{R} (N, X) \to {Hom}_{R} (M, X)$ , $g \mapsto g \circ u$ (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L5]

For a commutative ring $R$ and $R$ -modules $M, N$ , the abelian group ${Hom}_{R} (M, N)$ is an $R$ -module under $(r f) (m) = r f (m)$ (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L6]

For a commutative ring $R$ , $n \in N$ and an $R$ -module $N$ , the map $f \mapsto (f (e_{1}), \dots, f (e_{n}))$ is an isomorphism of $R$ -modules ${Hom}_{R} (R^{n}, N) \to N^{n}$ (For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$ ).

[L7]

Every finitely generated left module over a left Noetherian ring is Noetherian (Finitely generated modules over a left Noetherian ring are Noetherian).

[L8]

A finite direct sum is Noetherian if and only if every summand is Noetherian (Finite direct sums preserve and reflect Noetherian and Artinian conditions).

[L9]

In $R^{(X)}$ the standard basis vector $e_{x}$ has coordinate $1_{R}$ at $x$ and zero elsewhere, and every element is uniquely a finite $R$ -linear combination of them (The free module on a set and its standard basis).

[L10]

A left $R$ -module is Noetherian when every submodule of it is finitely generated (Noetherian modules: every submodule is finitely generated).

Proof

technique · direct

1.1

Fix a finite generating set $m_{1}, \dots, m_{n}$ of $M$ , with $n \in N$ . The universal property of the free module gives $π ⁣ : R^{n} \to M$ with $π (e_{i}) = m_{i}$ , and its image is the set of finite sums $\sum_{i} r_{i} m_{i}$ , that is $⟨ m_{1}, \dots, m_{n} ⟩_{R} = M$ ; so $π$ is surjective.

L1L2L3L9given

2.1

Precomposition with $π$ gives $π^{*} ⁣ : {Hom}_{R} (M, N) \to {Hom}_{R} (R^{n}, N)$ , $g \mapsto g \circ π$ . It is $R$ -linear for the module structures above, since $(g + g^{'}) \circ π = g \circ π + g^{'} \circ π$ and $(r g) \circ π = r (g \circ π)$ , both by evaluating at a point of $R^{n}$ . It is injective: if $g \circ π = 0$ then $g$ vanishes on $im π = M$ , so $g = 0$ .

L4L5step 1.1

2.2

The module ${Hom}_{R} (R^{n}, N)$ is Noetherian. Indeed $N$ is finitely generated over the Noetherian ring $R$ , hence a Noetherian module; the finite direct sum $N^{n}$ of copies of $N$ is then Noetherian; and ${Hom}_{R} (R^{n}, N)$ is isomorphic to $N^{n}$ , while an isomorphism of modules carries submodules to submodules and finite generating sets to finite generating sets, so the isomorphic module is Noetherian too.

L6L7L8L10step 1.1

3.1

The image $π^{*} ({Hom}_{R} (M, N))$ is a submodule of the Noetherian module ${Hom}_{R} (R^{n}, N)$ , hence finitely generated; and $π^{*}$ is injective and $R$ -linear, so ${Hom}_{R} (M, N)$ is isomorphic to that image and is therefore finitely generated as well.

L10step 2.1step 2.2∎

Remarks

The injectivity of $π^{*}$ is an instance of left exactness. Covariant and contravariant $Hom$ are left exact gives it for any exact $A \to B \to C \to 0$ ; step 2.1 writes out the case needed here, which uses only that $π$ is surjective and so does not require assembling the exact sequence first.
No claim of surjectivity. A homomorphism $R^{n} \to N$ descends to $M$ exactly when it kills $\ker π$ , and most do not; the corollary needs only the embedding.
Both hypotheses of finite generation are used, and for different reasons. Finite generation of $M$ produces the free cover in step 1.1; finite generation of $N$ makes the target Noetherian in step 2.2.

Depends on

Used by

Hom_ℤ(ℤ/m,ℤ/n)≅ℤ/gcd(m,n) for n≥1 Example

Dependency tree · two levels

23 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., Exercise (16.20) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §1 and §3 (standard reference, not scraped)