Is every commutative group structure underlying at least one (unitary, commutative) ring structure

Question

From the theorem of classification of finitely generated abelian groups, we can see that every finitely generated commutative group can be considered as the additive structure underlying (at least) one unitary and commutative ring. My question is about possible generalization of this result. Question: Is it true that with every abelian group G it is possible to build (at least) one ring(resp: unitary ring, commutative ring) whose additive structure is the group structure on G ? Gérard Lang

Answer 1

As Sasha Anan'in already mentioned, there are counterexamples like $\mathbb{Z}/p^\infty = \mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ and $\mathbb{Q}/\mathbb{Z}$. The common feature of these groups is that there are divisible and torsion and therefore $A\otimes_\mathbb{Z}A=0$. A ring $A$ has a surjective and hence non-zero map $A\otimes_\mathbb{Z}A\to A$ namely the multiplication. Therefore no additive group with $A\otimes_\mathbb{Z}A=0$ can be the additive group of a ring.

Answer 2

No, there are many abelian groups such that only zero multiplication can be defined over them. Even if you can define a non-zero multiplication on a group, the ring obtained may have no unit, also it may not be associative, and so on. A detailed discussion of this subject can be find in L. Fuchs: (Infinite abelian groups). Two former Ph.D students of our department have also some papers in the subject. You can find their works searching the names: "A. Najafizadeh, F. Karimi, A. M. Aghdam", and key words "torsion free groups, additive group of rings". One more interesting problem is to determine which subgroups of an abelina group $A$ can be realized as ideals in some ring with additive group $A$. You can find some good results concerning this problem.