 Solution

Put $D=\mathbb{Q}/\mathbb{Z}$. The character module $M^{\ast }={\mathrm{Hom}}_{\mathbb{Z}}(M,D)$ has the natural actionAdditivity is immediate, and $a(b\varphi )(m)=\varphi (bam)=\varphi (abm)$, using commutativity of $A$. The identity of $A$ acts as the identity. We establish the extension and duality tools used in the four implications, so no flatness or injectivity criterion is assumed without proof.

An injective module $E$ is one for which every homomorphism from a submodule $U\subseteq V$ extends to $V$. The Baer criterion says it suffices to test the inclusions of ideals $I\subseteq A$. Necessity follows directly from the definition. For sufficiency, suppose every $I\to E$ extends to $A$. Given $h:U\to E$, use Zorn's lemma to choose a maximal extension $(W,h_W)$ inside $V$; extensions along a chain have a well-defined union. If $v\in V\setminus W$, letThis is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Thenis well-defined: if two expressions represent the same element, their coefficient difference lies in $I$, where the two prescribed values agree. It extends $h_W$ to the strictly larger module $W+Av$, a contradiction. Thus $W=V$, proving the Baer criterion for arbitrary modules and ideals.

The group $D$ is divisible. A homomorphism $n\mathbb{Z}\to D$, for $n>0$, extends to $\mathbb{Z}$ by choosing $d\in D$ with $nd$ equal to its value at $n$; the zero ideal causes no restriction. The proved Baer criterion over $\mathbb{Z}$ therefore makes $D$ injective. It is also a cogenerator: a nonzero element $x$ of an abelian group generates either a finite cyclic group of order $n>1$, which maps $x$ to $1/n+\mathbb{Z}$, or an infinite cyclic group, which maps $x$ to $1/2+\mathbb{Z}$. Injectivity extends this character to the whole group. Thus every nonzero element is detected by an additive character.

It follows that a group homomorphism $u:T\to U$ is injective exactly whenis surjective. For an injection, extension into the injective group $D$ proves surjectivity. Conversely, a character nonzero on an element of $\ker u$ cannot be the pullback of a character on $U$. This is the detection property of the injective cogenerator of abelian groups.

Finally, for any $A$-module $N$ there is a natural character form of the Tensor-hom adjunction:An $A$-linear map $h$ gives the pairing $(n,m)\mapsto h(n)(m)$. It is balanced because $h(an)(m)=(ah(n))(m)=h(n)(am)$. Conversely a balanced additive pairing determines $h$ by this formula. The universal property of the tensor product of modules makes these constructions mutually inverse, and compatible with restriction maps.