# Is the class of additive groups of rings axiomatizable?

I know that it is impossible to axiomatize the multiplicative structures of rings, called $R$-semigroups. Is anything known about the first-order axiomatizability of the class of abelian groups which are additive groups of some ring? I don't want to restrict the meaning "ring" here. I would like to know whether this question is answered for rings with any subset of the set of adjectives {"associative", "unitary", "commutative"} attached.

EDIT I forgot to mention that I do want to exlude some rings, that is rings with zero multiplication.

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When a subset of your adjectives doesn't contain "unitary," the answer is trivial, because you can take any abelian group and make it a ring by defining all products to be zero. – Andreas Blass Mar 29 '12 at 12:11
... and every finitely generated abelian group is the additive group of some associative unitary commutative ring, by looking at the classification: en.wikipedia.org/wiki/…. Is there any (necessarily not finitely generated) abelian group which is not the underlying additive group of some commutative ring? – Sean Eberhard Mar 29 '12 at 12:26
@Sean: I assume you mean unitary rings. How about $\mathbb{Z}_2 + \mathbb{Z}_3+...+\mathbb{Z}_p+...$? Suppose that $\alpha$ is the generator of $\mathbb{Z}_2$, $\beta$ is the generator of $\mathbb{Z}_3$,... and $k\alpha+m\beta+...$ is the unit in the ring, $k$ is $0$ or $1$, $m$ is $0,1$ or $2$,... . Consider $\alpha*\beta$. Note that $2\alpha*\beta=3\alpha*\beta=0$, hence $\alpha*\beta=\beta*\alpha=0$. Also $(k\alpha+m\beta)*s=s$, $k\alpha*s+m\beta*s=s$ - for every $s$. Taking $s=\alpha$, we get $k\alpha*\alpha+m\beta*\alpha=\alpha$. Hence $k\alpha*\alpha=\alpha$. – Mark Sapir Mar 29 '12 at 13:11
The trivial group, with the only possible ring structure, is unitary, so by excluding zero multiplication you are in some sense excluding too much. – Tom Goodwillie Mar 29 '12 at 13:18
@Sean: Here is a torsion-free example. Consider a nontrivial subgroup of $\mathbb Q$. If it admits a nonzero multiplication, then it is isomorphic (scaling by some rational number) to a unitary subring of $\mathbb Q$. This excludes examples like the group of all rational numbers with square-free denominator. – Tom Goodwillie Mar 29 '12 at 13:26

In particular, an abelian group $A$ is elementarily equivalent to our favorite group $\mathbb Z$ iff $A$ is torsion free and of $p$-rank 1 for every prime $p$. The $p$-rank of $A$ is defined to be the minimum of $\textrm{dim}_{\mathbb Z_p}(A/pA)$ or $\omega$. An example of a torsion free abelian group of $p$-rank 1 for every $p$ is the subgroup $S\leq \mathbb Q$ consisting of rationals with square free denominator.
So $\mathbb Z$ and $S$ are elementarily equivalent. $\mathbb Z$ is the additive subgroup of the unital ring $\mathbb Z$, while (as Tom Goodwillie has pointed out) $S$ is not the additive subgroup of any unital ring. In fact, it is impossible to equip $S$ with any nonzero bi-additive multiplication. For, if $s, t\in S$, then $s$ and $t$ are $p$-divisible for almost all primes $p$. By bi-additivity, $s*t$ is $p^2$-divisible for almost all primes $p$. But the only element of $S$ that is $p^2$-divisible for almost all $p$ is 0, so $s*t=0$.