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

Statement

Let $R$ be a commutative ring and let $M, N$ be $R$ -modules. Then the abelian group ${Hom}_{R} (M, N)$ of The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition becomes an $R$ -module under the pointwise scalar action

$(r f) (m) : = r f (m) (r \in R, f \in {Hom}_{R} (M, N), m \in M) .$

The underlying additive group is the published one, unchanged: this extends the abelian-group structure rather than replacing it. Commutativity of $R$ is used, and is used only to see that $r f$ is again $R$ -linear.

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N$ .

[L1]

For left $R$ -modules $M, N$ , the set ${Hom}_{R} (M, N)$ of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse $(- f) (m) = - f (m)$ (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L2]

A ring $R$ is commutative when its multiplication is commutative, $x y = y x$ for all $x, y \in R$ (Commutative ring).

[L3]

A left $R$ -module is an abelian group $(M, +, 0_{M})$ with an action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L4]

A function $f : M \to N$ between left $R$ -modules is an $R$ -module homomorphism if $f (m + m^{'}) = f (m) + f (m^{'})$ and $f (r m) = r f (m)$ for all $m, m^{'} \in M$ and $r \in R$ (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

The set ${Hom}_{R} (M, N)$ already carries the pointwise addition making it an abelian group, with the zero homomorphism as neutral element and $(- f) (m) = - f (m)$ as inverse. Nothing below alters that addition.

L1given

2.1

For $r \in R$ and $f \in {Hom}_{R} (M, N)$ the function $r f$ defined by $(r f) (m) = r f (m)$ is again an $R$ -module homomorphism. Additivity is pointwise: $(r f) (m + m^{'}) = r (f (m) + f (m^{'})) = r f (m) + r f (m^{'})$ . Homogeneity is where commutativity enters: for $s \in R$ , $(r f) (s m) = r f (s m) = r (s f (m)) = (r s) f (m) = (s r) f (m) = s (r f (m)) = s (r f) (m)$ .

L2L3L4step 1.1algebra

3.1

The module axioms hold pointwise, each being an identity in $N$ evaluated at an arbitrary $m \in M$ : $((r + r^{'}) f) (m) = (r + r^{'}) f (m) = r f (m) + r^{'} f (m)$ , $(r (f + g)) (m) = r (f (m) + g (m)) = r f (m) + r g (m)$ , $((r r^{'}) f) (m) = (r r^{'}) f (m) = r (r^{'} f (m))$ and $(1_{R} f) (m) = 1_{R} f (m) = f (m)$ .

L1L3step 2.1algebra

4.1

So ${Hom}_{R} (M, N)$ with the addition of step 1.1 and the action of step 2.1 satisfies the definition of an $R$ -module, and its additive group is the published abelian group unchanged.

step 1.1step 3.1∎

Remarks

Commutativity is not decoration here. Over a noncommutative $R$ the function $r f$ need not be $R$ -linear, since the computation in step 2.1 turns on $r s = s r$ . That is why The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition gives only an abelian group in general, and why the module structure is recorded separately rather than being read into the definition.
The induced maps are $R$ -linear for this structure. For $u ⁣ : M \to M^{'}$ and $v ⁣ : N \to N^{'}$ the maps $u^{*}$ and $v_{*}$ of The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition satisfy $u^{*} (r g) = r u^{*} (g)$ and $v_{*} (r f) = r v_{*} (f)$ , both by evaluating at a point, so nothing about the published functoriality has to be revisited.

Depends on

Used by

Dependency tree · two levels

9 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., §4 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §1 (standard reference, not scraped)

Over a commutative ring the homomorphism group Hom⁡R(M,N)HomR​(M,N) is an RR-module