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Does every Abelian group admit a ring structure?

How might I construct an abelian group that cannot be represented by any ring with its addition operation? Wikipedia suggests 3 related topics, but they are quite unfamiliar. Perhaps it is possible to construct simple examples?

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    You won't find any finite, or even finitely generated examples, since those are all products of cyclic groups, which have natural interpretations as rings. – Gerry Myerson Feb 17 '12 at 00:02
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    Will you please specify what Wikipedia page you are referring to? This problem came up before at the question [Does every Abelian group admit a ring structure?](http://math.stackexchange.com/a/93411). – Jonas Meyer Feb 17 '12 at 00:02
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    I assume your definition of ring requires that it have a unity, correct? Otherwise any abelian group can be turned into a ring by making the product of any two elements be $0$. – Alex Becker Feb 17 '12 at 00:03
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    Have you seen [this question](http://math.stackexchange.com/questions/106933/showing-there-is-no-ring-whose-additive-group-is-isomorphic-to-mathbbq-math)? I don't think that $\mathbf Q/\mathbf Z$ is so exotic. – Dylan Moreland Feb 17 '12 at 00:03
  • @JonasMeyer: Sorry for the delay, there was something wrong with my internet connection. Here is the link http://en.wikipedia.org/wiki/Abelian_group#Additive_groups_of_rings – Fred Wilson Feb 17 '12 at 09:02
  • @FredWilson: No examples are given there, just the statement that the bulleted things are "important topics in this area of study." But there are examples given in the answers to the question linked in my first comment, including, if you restrict to unital rings, $\mathbb Z_2\oplus\mathbb Z_3\oplus\mathbb Z_4\oplus\cdots=\bigoplus\limits_{n=2}^\infty\mathbb Z_n$. Is your question answered by that? If so, it would make sense to close this one as duplicate, because we might as well collect the answers in one place instead of two. If not, will you please elaborate on how it is different? – Jonas Meyer Feb 17 '12 at 13:21
  • @JonasMeyer: oh, I misread, my bad. Thanks for pointing that out. I think you are very right, these are essentially the same question. Thanks also for the link! – Fred Wilson Feb 17 '12 at 18:11
  • @Fred: It is not defined. A direct *product* of groups (or other structures) has all of the sequences you could create with components from said groups, but a direct *sum* has only those sequences for which finitely many are nonidentity. (In these additive groups, $0$ is the identity.) – anon Feb 17 '12 at 19:57
  • @Fred: The direct *sum* of infinitely many groups (as distinguished from the direct product) is the subset of the Cartesian product in which all but finitely many of the components are the identity elements of the respective groups, with pointwise operations. For example, $\bigoplus\limits_{n=2}^\infty\mathbb Z_n=\{(a_n)_{n=2}^\infty:\forall n,a_n\in \mathbb Z_n\text{ and }\exists m,\forall n>m,a_n=0\}$. While this group cannot be the additive group of a ring with identity, it can be given its usual componentwise multiplication to make it a "rng". For $\mathbb Q/\mathbb Z$ and...... – Jonas Meyer Feb 17 '12 at 19:58
  • ...some other groups, there is no nonzero multiplication making them rings (even without identity), as indicated in a comment by Jack Schmidt in the other thread. – Jonas Meyer Feb 17 '12 at 19:59
  • @JonasMeyer: thank you, so wouldn't taking the direct sum of infinitely many $\mathbb Z_2$ do the same trick? Since the identity has to have finite number of nonzero entries – Fred Wilson Feb 17 '12 at 20:24
  • @FredWilson: It is more subtle. The goal is to give an example for which you can prove that it is impossible to define *any* multiplication that makes it a ring with identity. Although we might hope that $(1,1,1,\ldots)$ would be there, we only hoped for that because we were thinking about componentwise multiplication. And in fact, $\mathbb Z_2\oplus\mathbb Z_2\oplus\cdots$ *is* (isomorphic to) the abelian group of a ring with identity, namely the polynomial ring $\mathbb Z_2[x]$. With $\mathbb Z_2\oplus Z_3\oplus\cdots$ there is no such ring (see other question for reason). – Jonas Meyer Feb 17 '12 at 20:44
  • If $(n_j)_{j=1}^\infty$ is a sequence of positive integers, then $\mathbb Z_{n_1}\oplus \mathbb Z_{n_2}\oplus\mathbb Z_{n_3}\oplus\cdots$ is the abelian group of a ring with identity if and only if $(n_j)$ is bounded. – Jonas Meyer Feb 17 '12 at 20:56

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