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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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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
I realized I forgot to exclude zero multiplication right after I posted but I didn't have time to edit. Sorry! I'm editing now. –  Michał Masny Mar 29 '12 at 12:52
@SeanEberhard Yes, as far as I know there is only zero multiplication on every divisible torsion group. –  Michał Masny Mar 29 '12 at 13:08
@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

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