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Wikipedia article gives a number of definitions of injective modules, namely:

  1. 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 $K$
  2. Any short exact sequence $0 \rightarrow Q \rightarrow M \rightarrow K \rightarrow 0$ of left $R$-modules splits
  3. 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 $hf = g$
  4. 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.

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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).

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  • $\begingroup$ Once we have an injection $Q\to Q'$ with $Q'$ injective, then by (2) $Q'\cong Q\oplus Q'/Q$. This implies that $Q$ is injective, then 3) follows. Of course we need the fact that $Q\cong M\oplus N$ is injective iff $M$ and $N$ is injective. $\endgroup$ – user114539 Aug 29 '16 at 11:50
  • $\begingroup$ @user114539 Which requires proving that a direct summand of an injective module is injective (basically what I do above). $\endgroup$ – egreg Aug 29 '16 at 16:44
  • $\begingroup$ True. Somehow I dont like this proof. If you see a book like Dummit and Foote, they give 3 equivalent properties, ask the reader to prove it and then define an injective module as one satisfying one and hence all the properties. But while proving the direct summand property ((2) implies the lifting property((3)), we are using a definition of injective modules as it being that of lifting property. But actually that definition was not given before beginning such a theorem. I dont know if this makes sense, perhaps some chat or direct conversation is necessary to convey what i mean. $\endgroup$ – user114539 Aug 31 '16 at 14:04
  • $\begingroup$ @user114539 I don't know Dummit and Foote; however, the technique I use for proving that if $Q$ splits in every extension then it is injective basically proves that a direct summand of an injective module is injective. So I don't see much of a simplification in what you propose. It might be a lemma of its own, of course. $\endgroup$ – egreg Aug 31 '16 at 15:00

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