Wikipedia article gives a number of definitions of injective modules, namely:
- If $Q$ is a submodule of some other left $R$-module $M$, then there exists another submodule $K$ of $M$ such that $M$ is the internal direct sum of $Q$ and $K$
- Any short exact sequence $0 \rightarrow Q \rightarrow M \rightarrow K \rightarrow 0$ of left $R$-modules splits
- If $X$ and $Y$ are left $R$-modules and $f : X \rightarrow Y$ is an injective module homomorphism and $g : X \rightarrow Q$ is an arbitrary module homomorphism, then there exists a module homomorphism $h : Y \rightarrow Q$ such that $hf = g$
- The contravariant functor $Hom(-,Q)$ from the category of left $R$-modules to the category of abelian groups is exact.
Where can I see a proof of equivalence of these four? I tried to google a bit, but didn't find a full and understandable proof. Alternatively, I would be very grateful if someone would post a proof here as an answer.
Best Answer
Theorem. Every (right) module is a submodule of an injective module.
The proof of this will be given later on, for the moment assume it. Using right modules is easier, but the result of course holds also for left modules.
Conditions 1 and 2 are clearly equivalent, as are conditions 3 and 4.
Suppose 2 holds. Let $Q'$ be an injective module such that $Q\subseteq Q'$. Let $j$ be the inclusion map. By the injectivity of $Q'$, the homomorphism $ig\colon X\to Q'$ can be extended to a homomorphism $h'\colon Y\to Q'$ such that $h'f=jg$. Since the inclusion $j$ splits, let $q\colon Q'\to Q$ be such that $qj=1_Q$ (the identity on $Q$). Then $$ qh'f=qjg=g $$ and $h=qh'\colon Y\to Q$ is the homomorphism we were looking for.
Assume 3 holds and let $Q$ be a submodule of $M$. Then the inclusion map $j\colon Q\to M$ admits an extension $q\colon M\to M$ such that $qj=1_Q$, so $Q$ is a direct summand of $M$.
Therefore 2 and 3 are equivalent and we are done.
The existence of an embedding $M\to E$ where $E$ is injective can be proved in the following way. The module $M$ embeds into $\operatorname{Hom}_{\mathbb{Z}}(R,M)$. If $M$, considered as an abelian group, is embedded in the abelian group $N$, then $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ is embedded in $\operatorname{Hom}_{\mathbb{Z}}(R,N)$, as it's easy to show. So we just need two lemmas.
Lemma. If $N$ is a divisible group, then $\operatorname{Hom}_{\mathbb{Z}}(R,N)$ is an injective $R$-module.
Proof. Let $f\colon X\to Y$ be an injective homomorphism and suppose $g\colon X\to \operatorname{Hom}_{\mathbb{Z}}(R,N)$ is a module homomorphism. Then we can consider $g'\colon X\to N$ defined by $g'(x)=g(x)(1)$. This is a group homomorphism, so by the divisibility of $N$, there is a group homomorphism $h'\colon Y\to N$ such that $h'f=g'$. Now define $h\colon Y\to\operatorname{Hom}_{\mathbb{Z}}(R,N)$ by $h(y)(r)=h'(yr)$.
Then, for $x\in X$ and $r\in R$, $$ hf(x)(r)=h(f(x))(r)=h'(f(x)r)=h'(f(xr))=g'(xr)=g(xr)(1)=g(x)(r) $$ so $hf=g$. QED
Note: the $R$-module structure on $\operatorname{Hom}_{\mathbb{Z}}(R,N)$ is given by defining $fr$ to be the map $fr\colon s\mapsto f(rs)$.
Lemma. Every abelian group embeds in a divisible group.
Proof. If $G$ is an abelian group, it is not restrictive to see $G=F/K$, where $F$ is a free abelian group. A free abelian group is a direct sum of copies of $\mathbb{Z}$, that so embeds in a direct sum $F'$ of copies of $\mathbb{Q}$, which is divisible. Then $G$ embeds in $F'/K$, which is also divisible. QED
Finally, notice that for abelian groups (or $\mathbb{Z}$-modules), being injective is equivalent to being divisible (apply Baer's criterion).