= Solution
Put $D=\mathbb Q/\mathbb Z$. The <character module> $M^*=\operatorname{Hom}_{\mathbb Z}(M,D)$ has the natural action
$$
\boxed{(a\phi)(m)=\phi(am).}
$$
Additivity is immediate, and $a(b\phi)(m)=\phi(bam)=\phi(abm)$, 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 $v\in V\setminus W$, let
$$
I=\{a\in A:av\in W\},\qquad \theta(a)=h_W(av).
$$
This is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Then
$$
h'(w+av)=h_W(w)+ae
$$
is 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 \b[every nonzero element is detected by an additive character].
It follows that a group homomorphism $u:T\to U$ is injective exactly when
$$
u^*: \operatorname{Hom}_{\mathbb Z}(U,D)
\longrightarrow\operatorname{Hom}_{\mathbb Z}(T,D)
$$
is 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>:
$$
\boxed{\operatorname{Hom}_A(N,M^*)
\simeq\operatorname{Hom}_{\mathbb Z}(N\otimes_AM,D).}
$$
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.
Back to article page
