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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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    $\begingroup$ 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. $\endgroup$
    – Andreas Blass
    Commented Mar 29, 2012 at 12:11
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    $\begingroup$ ... 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? $\endgroup$
    – Sean Eberhard
    Commented Mar 29, 2012 at 12:26
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    $\begingroup$ @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 $k\alpha*\alpha=\alpha$. $\endgroup$
    – user6976
    Commented Mar 29, 2012 at 13:11
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    $\begingroup$ The trivial group, with the only possible ring structure, is unitary, so by excluding zero multiplication you are in some sense excluding too much. $\endgroup$
    – Tom Goodwillie
    Commented Mar 29, 2012 at 13:18
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    $\begingroup$ @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. $\endgroup$
    – Tom Goodwillie
    Commented Mar 29, 2012 at 13:26

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The answer to the question in the title is No. You can prove this from the work of Wanda Szmielew on the elementary properties of abelian groups. This answer works for any kind of nonzero, bi-additive, binary multiplication (associative or not, commutative or not, unital or not).

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 $s*t=0$.

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