What is an injective module over a ring \(R\)? State Baer’s Criterion and the characterization of injective modules over PIDs (divisible modules).
Solution
A left \(R\)-module \(Q\) is injective if every homomorphism into \(Q\) defined on a submodule extends across the containing module: whenever \(A\hookrightarrow B\) is injective and \(f:A\to Q\) is \(R\)-linear, there exists \(g:B\to Q\) with \(g|_A=f\). Equivalently, the contravariant functor \[ \operatorname{Hom}_R(-,Q) \] is exact.
Baer’s criterion. A left \(R\)-module \(Q\) is injective if and only if every \(R\)-linear map \[ f:I\to Q \] from a left ideal \(I\subseteq R\) extends to an \(R\)-linear map \(R\to Q\).
Now let \(R\) be a commutative PID. An \(R\)-module \(Q\) is divisible if, for every \(0\ne r\in R\) and \(q\in Q\), there exists \(y\in Q\) with \[ ry=q. \] Over a PID, \[ Q\text{ is injective}\iff Q\text{ is divisible}. \] Indeed, every ideal is \((r)\). A homomorphism \(f:(r)\to Q\) is determined by \(q=f(r)\). Extending \(f\) to \(R\) is equivalent to finding \(y=g(1)\) with \(ry=q\), exactly the divisibility condition.
Thus over \(\mathbb Z\) the injective modules are precisely the divisible abelian groups; examples include \(\mathbb Q\), \(\mathbb Q/\mathbb Z\), and the Prüfer \(p\)-groups. More generally, if \(R\) is a PID, its fraction field \(\operatorname{Frac}(R)\) is divisible and therefore injective as an \(R\)-module.