All rings below are commutative with unity. For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.

Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $\hat {U(R)} := Hom_{\mathbb Z} (U(R),\mathbb Z)$.

Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) \cong \hat {U(R)}$ ?

In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)\cong Hom_{\mathbb Z} (G,\mathbb Z)$, for some ring $S$ ?

My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(\mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.

I don't know what happens if the group has a torsion.

Are there any other general sufficient or necessary conditions ?

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    $\begingroup$ Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$." $\endgroup$ – dan_fulea Jan 31 '19 at 18:05

The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $\mathbb{Z}$ is torsion-free: if $f\in\operatorname{Hom}(G,\mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $\mathbb{Z}$ for all $x\in G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.

  • $\begingroup$ Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$ $\endgroup$ – user640299 Feb 1 '19 at 13:53

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