# On the dual group of the group of units of commutative ring

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 ?

• 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 $U(S)=G$." – dan_fulea Jan 31 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.
• Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$ – user640299 Feb 1 at 13:53