Is every abelian group the additive group of some ring?

I would very much appreciate if someone could show me if this is false or true, something I'm thinking about and finding hard to prove so im starting to think its false. Thanks

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    depends on what you mean by ring. If your rings don't have to be unital, then for every abelian group, you can just define multiplication as $a \cdot b = 0$. – Dustan Levenstein Feb 21 '14 at 14:48
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    For unital rings it's at least true for finite abelian groups. These are all isomorphic to products of $\mathbb Z/n\mathbb Z$ and $\mathbb Z/n\mathbb Z$ has a natural ring structure. I'm not sure about infinite groups though. – Jim Feb 21 '14 at 14:53
  • It will also be true for finitely generated abelian groups because of the existence of the structure theorem. Any finitely generated abelian group is a a product of some $\mathbb{Z} / n_i \mathbb{Z}$ and finitely many copies of $\mathbb{Z}$ itself. As all of these factors are rings, the product can be given the product ring structure. – Singhal Feb 21 '14 at 15:08

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