Divisible torsion $\mathbb{Z}$-modules

Question

I am trying to prove that for any divisible torsion $\mathbb{Z}$-module $V$, this map $$f:\mathbb{Q}/\mathbb{Z}\otimes_E\text{Hom}(\mathbb{Q}/\mathbb{Z},V)\longrightarrow V\mbox{ defined by } f((q+\mathbb{Z})\otimes g)=g(q+\mathbb{Z})$$ is an isomorphism, where tensor is taken over the ring $E=\operatorname{End}(\mathbb Q/\mathbb Z)$.

It is easy to prove that $f$ is a homomorphism, but I couldn't prove that $f$ is bijective. Are there any special properties for divisible torsion $\mathbb{Z}$-modules that help in proving that the above map $f$ is bijective?

Answer 1

First, consider an element $v\in V$. As $V$ is torsion we can choose $n$ such that $n!v=0$. For $k\leq n$ put $u_k=(n!/k!)v$. Then choose $u_k$ for $k>n$ inductively with $ku_k=u_{k-1}$ (which is possible because $V$ is divisible). There is then a unique homomorphism $\phi\colon\mathbb{Q}/\mathbb{Z}\to V$ such that $\phi([1/k!])=u_k$ for all $k$, and $f([1/n!]\otimes \phi)=v$. This shows that $f$ is surjective.

Now consider an element $\alpha\in\ker(f)$. As $\mathbb{Q}/\mathbb{Z}$ is the union of the cyclic subgroups generated by elements of the form $[1/n!]$, we see that $\alpha$ can be written as $[1/n!]\otimes\phi$ for some $\phi\colon\mathbb{Q}/\mathbb{Z}\to V$ with $\phi([1/n!])=0$. Now multiplication by $n!$ gives a surjective endomorphism of $\mathbb{Q}/\mathbb{Z}$, whose kernel is generated by $[1/n!]$. It follows easily that $\phi=n!\psi$ for some $\psi$, and thus that $\alpha=n![1/n!]\otimes\psi=0$. Thus, $f$ is also injective.

The above argument shows that the composite $$ \mathbb{Q}/\mathbb{Z} \otimes_{\mathbb{Z}} \text{Hom}(\mathbb{Q}/\mathbb{Z},V) \to \mathbb{Q}/\mathbb{Z} \otimes_{\text{End}(\mathbb{Q}/\mathbb{Z})} \text{Hom}(\mathbb{Q}/\mathbb{Z},V) \to V $$ is an isomorphism, and it follows easily that both of the maps involved are isomorphisms. One can also see more directly that the first map is an isomorphism, using the standard fact that $\text{End}(\mathbb{Q}/\mathbb{Z})$ is the profinite completion of the integers, together with the obvious fact that the first tensor factor $\mathbb{Q}/\mathbb{Z}$ is a torsion group.

Answer 2

Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathbf{Z},V)/(n) \rightarrow V[n]$ defined by evaluation at $1/n \bmod \mathbf{Z}$ is an isomorphism (as then with a small diagram chase we can pass to the direct limit over more divisible $n$ to conclude).

But $V$ is an injective abelian group (as $V$ is divisible), so applying ${\rm{Hom}}(\cdot, V)$ to the exact sequence $$0 \rightarrow (1/n)\mathbf{Z}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \stackrel{n}{\rightarrow} \mathbf{Q}/\mathbf{Z} \rightarrow 0$$ yields exactly the desired isomorphism.

[In effect, this is a reformulation of the same argument as in Neil Strickland's answer, as I am masking some calculations implicit in the proof that divisible abelian groups are injective. I am also tacitly using that $\widehat{\mathbf{Z}}$ is the endomorphism ring of $\mathbf{Q}/\mathbf{Z}$, so tensoring over it against a torsion module is the same as tensoring over $\mathbf{Z}$.]