TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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For a commutative ring $R$ , $R$ -linear $G$ -actions are exactly the compatible left $R [G]$ -module structures

Statement

Let $R$ be a commutative ring, let $G$ be a group, and let $M$ be a left $R$ -module. Then:

every $R$ -linear action of $G$ on $M$ extends uniquely to a left $R [G]$ -module structure on $M$ satisfying $(r [e]) \cdot m = r m (r \in R, m \in M);$
every left $R [G]$ -module structure on $M$ satisfying $(r [e]) \cdot m = r m (r \in R, m \in M)$ restricts to an $R$ -linear action of $G$ on $M$ ;
these two constructions are inverse to each other.

Moreover, if $M$ and $N$ carry the corresponding structures, then an $R$ -linear map $f : M \to N$ is $G$ -equivariant if and only if it is an $R [G]$ -module homomorphism.

Facts & Assumptions

Given: A commutative ring $R$ , a group $G$ , and left $R$ -modules $M$ and $N$ .

[L1]

The group ring $R [G]$ has basis vectors $[g]$ , multiplication $[g] [h] = [g h]$ , identity $[e]$ , and central scalar copy $r \mapsto r [e]$ (The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ , The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements).

[L2]

An $R$ -linear $G$ -module is a left $R$ -module together with a group action whose each $g$ -operator is $R$ -linear (An $R$ -linear action of $G$ on a left $R$ -module, and a $G$ -module over $R$ ).

[L3]

Every set map from $G$ to a left $R$ -module extends uniquely to an $R$ -module homomorphism from $R [G]$ (Universal property of the free module on a set).

[L4]

An $R [G]$ -module homomorphism is a map compatible with addition and with the scalar action of every element of $R [G]$ (Module homomorphism and isomorphism, kernel, image and cokernel, Unital left and right modules over a ring; unqualified module means left module).

Proof

technique · constructive

1.1L1L2L3givenconstruct

Assume first that $M$ is an $R$ -linear $G$ -module. For each $m \in M$ , the map $u_{m} : G \to M$ , $u_{m} (g) : = g \cdot m$ , extends uniquely by [L3] to an $R$ -linear map ${\tilde{u}}_{m} : R [G] \to M$ . Define $a \cdot m : = {\tilde{u}}_{m} (a)$ for $a \in R [G]$ .

2.1step 1.1L1L2

For basis elements, $[g] \cdot m = g \cdot m$ by construction. If $a = \sum_{g \in F} r_{g} [g]$ , then $a \cdot m = \sum_{g \in F} r_{g} (g \cdot m)$ , so the action is additive in $a$ and, because each $g$ -operator is $R$ -linear by [L2], also additive in $m$ and compatible with the scalar action of $R$ on $M$ .

3.1step 2.1L1L2givenalgebra

For $g, h \in G$ , one has $([g] [h]) \cdot m = [g h] \cdot m = (g h) \cdot m = g \cdot (h \cdot m) = [g] \cdot ([h] \cdot m)$ . Since both sides are $R$ -bilinear in the two group-ring variables, the equality extends to $(a b) \cdot m = a \cdot (b \cdot m)$ for all $a, b \in R [G]$ . Also $[e] \cdot m = e \cdot m = m$ , so the identity of $R [G]$ acts as the identity on $M$ , and $(r [e]) \cdot m = r (e \cdot m) = r m$ . Thus the construction of steps 1.1-2.1 gives a compatible left $R [G]$ -module structure on $M$ .

4.1step 3.1L1L2L4givenalgebra

Conversely, assume $M$ is a left $R [G]$ -module compatible with the given $R$ -module structure, meaning that $(r [e]) \cdot m = r m$ for all $r \in R$ and $m \in M$ . Define $g \cdot m : = [g] \cdot m$ . Then $e \cdot m = [e] \cdot m = m$ , and $(g h) \cdot m = [g h] \cdot m = ([g] [h]) \cdot m = [g] \cdot ([h] \cdot m) = g \cdot (h \cdot m)$ , so this is a group action. For $r \in R$ , one has $g \cdot (r m) = [g] \cdot ((r [e]) \cdot m) = (([g] (r [e])) \cdot m) = (((r [e]) [g]) \cdot m) = (r [e]) \cdot ([g] \cdot m) = r (g \cdot m)$ . Thus the restricted action is $R$ -linear.

5.1step 2.1step 4.1L1L4givenalgebra

Let $f : M \to N$ be $R$ -linear. If $f$ is an $R [G]$ -module homomorphism, then $f (g \cdot m) = f ([g] \cdot m) = [g] \cdot f (m) = g \cdot f (m)$ , so $f$ is $G$ -equivariant. Conversely, if $f$ is $G$ -equivariant and $a = \sum_{g \in F} r_{g} [g]$ , then $f (a \cdot m) = \sum_{g \in F} r_{g} f (g \cdot m) = \sum_{g \in F} r_{g} (g \cdot f (m)) = a \cdot f (m)$ , so $f$ is an $R [G]$ -module homomorphism.

6.1step 2.1step 4.1step 5.1L1discharge-construct∎

In the first construction, the recovered action of a basis element $[g]$ is the original action of $g$ by step 2.1, so the recovered $G$ -action is unchanged. In the second construction, the recovered $R [G]$ -action agrees with the original one on each basis element $[g]$ , and it has the same scalar action because the compatibility condition fixes $(r [e]) \cdot m = r m$ . Therefore the two constructions are inverse.

Depends on

Dependency tree · 2 levels

Sources

Peter Webb, A Course in Finite Group Representation Theory, Proposition 1.1.5

For a commutative ring RR, RR-linear GG-actions are exactly the compatible left R[G]R[G]-module structures