LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$

Statement

Let $R$ be a commutative ring, let $n \in N$ , let $R^{n}$ be the free $R$ -module on an $n$ -element set with standard basis $e_{1}, \dots, e_{n}$ (The free module on a set and its standard basis), and let $N$ be an $R$ -module. Write $N^{n}$ for the direct sum of $n$ copies of $N$ (The direct sum of an indexed family of modules), which for a finite index set is the coordinatewise product. Then

$Φ ⁣ : {Hom}_{R} (R^{n}, N) ⟶ N^{n}, Φ (f) = (f (e_{1}), \dots, f (e_{n})),$

is an isomorphism of $R$ -modules, the source carrying the module structure of Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module.

At $n = 0$ both sides are the zero module.

Facts & Assumptions

Given: A commutative ring $R$ , a natural number $n$ , the free module $R^{n}$ with standard basis $e_{1}, \dots, e_{n}$ , and an $R$ -module $N$ .

[L1]

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

[L2]

In $R^{(X)}$ the standard basis vector $e_{x}$ has coordinate $1_{R}$ at $x$ and zero elsewhere, and every element has a unique expression $\sum_{x \in F} r_{x} e_{x}$ with $F \subseteq X$ finite; for $X = \emptyset$ the module is $0$ (The free module on a set and its standard basis).

[L3]

For a family $(M_{i})_{i \in I}$ of left $R$ -modules the direct sum is the submodule of the coordinatewise product consisting of the families of finite support; for $I = \emptyset$ both product and direct sum are the zero module (The direct sum of an indexed family of modules).

[L4]

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

Proof

technique · direct

1.1L1L2given

$Φ$ is a bijection. It is injective: two homomorphisms $R^{n} \to N$ agreeing on $e_{1}, \dots, e_{n}$ are the unique extension of the same set map on the index set, hence equal. It is surjective: given $(y_{1}, \dots, y_{n}) \in N^{n}$ , the set map $i \mapsto y_{i}$ extends to a homomorphism $f ⁣ : R^{n} \to N$ with $f (e_{i}) = y_{i}$ , so $Φ (f) = (y_{1}, \dots, y_{n})$ .

2.1L3L4step 1.1algebra

$Φ$ is $R$ -linear. Addition in ${Hom}_{R} (R^{n}, N)$ and in $N^{n}$ is pointwise and coordinatewise respectively, so $Φ (f + g) = ((f + g) (e_{1}), \dots) = Φ (f) + Φ (g)$ ; and the scalar action on the source is pointwise, so $Φ (r f) = (r f (e_{1}), \dots, r f (e_{n})) = r Φ (f)$ .

3.1L2L3step 1.1step 2.1∎

A bijective $R$ -module homomorphism is an isomorphism of $R$ -modules, so $Φ$ is one. At $n = 0$ the index set is empty: $R^{0} = 0$ , the only homomorphism $0 \to N$ is the zero map, and $N^{0}$ is the zero module, so both sides are zero and $Φ$ is the unique map between them.

Remarks

The isomorphism depends on the chosen basis. A different ordered basis of $R^{n}$ gives a different $Φ$ ; what is canonical is that ${Hom}_{R} (R^{n}, N)$ is isomorphic to $N^{n}$ , not any particular isomorphism.
Finiteness of the index set is what makes the target a direct sum. For an infinite index set $X$ the same argument identifies ${Hom}_{R} (R^{(X)}, N)$ with the coordinatewise product of copies of $N$ , not with the direct sum, because a homomorphism may be nonzero on infinitely many basis vectors.

For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a commutative ring, Hom⁡R(Rn,N)≅NnHomR​(Rn,N)≅Nn

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a commutative ring, ${Hom}_{R} (R^{n}, N) ≅ N^{n}$