Proposition 13, Section 4.3 of Hungerford’s Algebra

Question

Proposition 3.12. Every unitary module $A$ over a ring $R$ with identity may be embedded in an injective $R$-module.

Proposition 3.7. A direct product of $R$-modules $\prod_{i\in I} J_i$ is injective if and only if $J_i$ is injective for every $i \in I$.

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Let R be a ring with identity. The following conditions on a unitary $R$-module $J$ are equivalent.

(i) $J$ is injective;

(ii) every short exact sequence $0\to J\to B \to C \to 0$ is split exact (hence $B\cong J\oplus C$);

(iii) $J$ is a direct summand of any module $B$ of which it is a submodule.

Sketch of proof: (ii) $\Rightarrow$ (iii) since the sequence $0\to J \xrightarrow{\subset} B \xrightarrow{\pi} B/J \to 0$ is split exact …

(iii) $\Rightarrow$ (i) It follows from Proposition 3.12 that $J$ is a submodule of an injective module $Q$. Proposition 3.7 and (iii) imply that $J$ is injective.

In proof of (iii) $\Rightarrow$ (i), I don’t understand “Proposition 3.7 and (iii) imply that $J$ is injective” sentence. By proposition 3.12, there exists an monomorphism $f:J\to Q$. So we can identify $J$ as submodule of injective module $Q$. By (iii), $\exists$ module $B$ such that $J\oplus C=B$ as internal direct sum.

In proof of (ii) $\Rightarrow$ (iii), $B$ could be $J$, right?

Answer 1

• $(3)\implies (1)$: by proposition $3.12$ $J$ is a submodule of an injective module $Q$. By $(3)$ there exists a submodule $I$ of $Q$ such that $Q=I\oplus J=I\times J$ (recall that finite sum and finite product are isomorhic). Since $Q$ is injective, it follows from proposition $3.7$ that also $J$ is injective ($J$ is in a product which gives $Q$).
• $(2)\implies (3)$: yes, it is possible that $B=J$. Note that in that case the statement is trivial since $J=J\oplus 0$.