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Coinduction sends injective abelian groups to injective modules

Statement

Let $R$ be a unital ring and $D$ an injective abelian group. Give ${Hom}_{Z} (R, D)$ the left $R$ -action $(r φ) (s) = φ (s r) .$ Then ${Hom}_{Z} (R, D)$ is an injective left $R$ -module.

Facts & Assumptions

Given: A unital ring $R$ and an injective abelian group $D$ .

[F1]

Injectivity is extension of homomorphisms along monomorphisms (Injective modules and the extension property).

[F2]

$Hom$ groups use pointwise addition and maps induced by composition (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

Proof

technique · direct

1.1givenF2algebra

The formula $(r φ) (s) = φ (s r)$ satisfies $r_{1} (r_{2} φ) = (r_{1} r_{2}) φ$ , the distributive laws, and $1 φ = φ$ , so it defines a left $R$ -module structure.

2.1step 1.1F2

For every left $R$ -module $M$ , evaluation at $1$ defines $Θ_{M} : {Hom}_{R} (M, {Hom}_{Z} (R, D)) \to {Hom}_{Z} (M, D), Θ_{M} (F) (m) = F (m) (1) .$

3.1step 1.1step 2.1algebra

The inverse sends $g : M \to D$ to $\hat{g} (m) (r) = g (r m)$ . Indeed, $\hat{g}$ is $R$ -linear because $\hat{g} (s m) (r) = g (r s m) = (s \hat{g} (m)) (r)$ , and evaluation at $1$ and the unit law show that the two constructions are inverse.

3.2givenstep 2.1F1

Let $u : A \to B$ be a monomorphism and $F : A \to {Hom}_{Z} (R, D)$ . Under $Θ_{A}$ , it corresponds to a group homomorphism $g : A \to D$ , which extends along the underlying subgroup inclusion to $\tilde{g} : B \to D$ because $D$ is injective.

4.1step 3.1step 3.2F1∎

The inverse construction of step 3.1 turns $\tilde{g}$ into an $R$ -linear extension $\hat{\tilde{g}} : B \to {Hom}_{Z} (R, D)$ of $F$ . Hence the coinduced module is injective by [F1].

Coinduction sends injective abelian groups to injective modules

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources