Without using Baer’s criterion show $\Bbb{Z}/n\Bbb{Z}$ is self-injective

Question

Unfortunately I have only seen examples of this using Baer’s criterion. I am following Wiebel’s an introduction to Homological Algebra as a text. I was given a hint:

Show that a morphism $f: A \to B $ between $\Bbb{Z}$-modules is injective iff the corresponding morphism $\text{Hom}_{\Bbb{Z}}(B, \Bbb{Q}/\Bbb{Z}) \to \text{Hom}_{\Bbb{Z}}(A, \Bbb{Q}/\Bbb{Z})$ is surjective. Which I have done, but I am unsure how this applies!

Thanks for any help!

Answer 1

Consider an injective morphism of $Z/n\mathbb Z$-modules $A \to B$ and a morphism $A \to \mathbb Z/n\mathbb Z$. Of course these are also morphisms of abelian groups.

Look at the diagram

$$\require{AMScd} \begin{CD} A @>>> B\\ @VVV \\ \mathbb Z/n\mathbb Z \\ @VV{\cdot \frac{1}{n}}V \\ \mathbb Q/\mathbb Z \end{CD}$$

By what you have shown, there is a map $B \to \mathbb Q/\mathbb Z$ making the diagram commutative. The fact that $B$ is actually a $\mathbb Z/n\mathbb Z$-module yields that this map factors over $\mathbb Z/n\mathbb Z$ and this is all you need.

Further note: Of course this is a diagram of morphism of abelian groups, but the upper triangle of the diagram consists of $\mathbb Z/n\mathbb Z$-modules and any morphism of abelian groups between $\mathbb Z/n\mathbb Z$-modules is also an $\mathbb Z/n\mathbb Z$-module morphism.

Answer 2

More generally, $\operatorname{Hom}_{\Bbb Z}(F,\Bbb Q/\Bbb Z)$ is an injective $R$-module for every ring $R$ and every flat right $R$-module $F$.

This is because if we consider $F$ as a $(\Bbb Z,R)$-bimodule $$\operatorname{Hom}_{\Bbb Z}(F,-):_{\Bbb Z}\mathbf{Mod}\to _{R}\mathbf{Mod}$$ is right adjoint to $$F\otimes_R-:_R\mathbf{Mod}\to _{\Bbb Z}\mathbf{Mod}$$ the latter functor is exact as $F$ is flat, thus it preserves monomorphisms.

Now it's an easy exercise in abstract nonsense that a functor which has a left adjoint that preserves monomorphisms maps injective objects to injective objects. If we apply this to our situation, we see that $\operatorname{Hom}_{\Bbb Z}(F,\Bbb Q/\Bbb Z)$ is injective as $\Bbb Q/\Bbb Z$ is an injective $\Bbb Z$-module. (This follows from what you have shown.)

If we set $R= \Bbb Z/n\Bbb Z$, then $R$ as a module over itself is free, hence flat, thus $\operatorname{Hom}_{\Bbb Z}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$ is injective, but $\operatorname{Hom}_{\Bbb Z}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z) \cong \Bbb Z/n\Bbb Z$.