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The endomorphisms of an abelian group form a ring under pointwise group operation and composition. Every ring is isomorphic to a subring of the endomorphism ring of some abelian group (left module over itself).

Is every ring isomorphic to the endomorphism ring of some abelian group? (not just a subring)

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Any endomorphism ring which has characteristic $p$ must be the endomorphism ring of an $\mathbb{F}_p$ vector space. But there are plenty of characteristic $p$ rings which are not of this form, for instance $\mathbb{F}_{p^k}$, $k>2$. – Kevin Ventullo Jul 17 '10 at 7:26
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Generally it's difficult to characterize rings isomorphic to an endomorphism ring of an abelian group. Interest in such problems was sparked by a problem given by Fuchs in his widely-read monograph Abelian Groups, cf. the excerpt below from the introduction to the paper [1]

The notion of an E-ring goes back to a seminal paper of Schultz [20] written in response to Problem 45 in the well-known book `Abelian Groups' by Laszlo Fuchs [11]. In this paper Schultz distinguished between two possibly different approaches, the first we will continue to call an E-ring, while the second we shall refer to as a generalized E-ring. Thus a ring R is said to be an E-ring if R is isomorphic to the endomorphism ring of its underlying additive group, R+, via the mapping sending an element r $\in$ R to the endomorphism given by left multiplication by r, whilst R is a generalized E-ring if some isomorphism, not necessarily left multiplication, exists between R and its endomorphism ring End(R+). Since right multiplication is always an endomorphism, it is not difficult to see that E-rings are necessarily commutative. The existence of a non-commutative generalized E-ring has recently been established [15], and so it follows that the class of generalized E-rings is strictly larger than the class of E-rings.

Since Schultz's original paper there has been a great deal of interest in E-rings and some natural generalizations, see e.g. [1,2,4,6,8-10,17,19,21]. A notable feature of much of this recent work has been the use of so-called realization theorems, whereby a cotorsion-free ring is realized, using combinatorial ideas derived from Shelah's Black Box - see e.g. [7] for details of this technique - as the endomorphism ring of an Abelian group. This present work arose from an observation of the second author in response to a question from the first about the existence of generalized E-algebras over the ring $J_p$ of p-adic integers; see [16] for further details. A natural question which arises, is to what extent is it necessary for a ring to be cotorsion-free in order to be a generalized E-ring and the principal objective of this work is to characterize generalized E-rings `modulo cotorsion-free groups.' The characterization is quite elementary but seems to have been overlooked heretofore. It should be noted that Bowshell and Schultz showed in [2] that a reduced cotorsion E-ring has the form $\prod_{p \in U} {\mathbb > Z}(p^{k_p}) \oplus \prod_{p\in V} J_p$ where $U,V$ are disjoint sets of primes.

1 R. Gobel, B. Goldsmith.
Classifying E-algebras over Dedekind domains
Jnl. Algebra, Vol. 306, 2006, 566-575

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A ring can be the endomorphism ring of an abelian group without being the endomorphism ring of its underlying additive group. – Qiaochu Yuan Jul 17 '10 at 19:07
It took me more than a few minutes to correctly parse this sentence: "its additive group" refers to the ring, not to the abelian group. To paraphrase Bill Clinton, whether or not this is correct depends on what "it" is. – Victor Protsak Jul 17 '10 at 21:07
My intent was to point out the the historical genesis of such problems and to give an example of some nontrivial results. This helps provide an entry-point into the literature. – Bill Dubuque Jul 17 '10 at 21:09
I edited my remark to help avoid any further confusion. – Bill Dubuque Jul 17 '10 at 22:19
My memory was correct, Fuchs did specify the general problem, namely "Problem 44. Find necessary and sufficient conditions for a ring (group) to be an endomorphism ring (group) of some group" in his Abelian Groups 1960, according to Google Books. Hence the genesis of the problem. – Bill Dubuque Jul 17 '10 at 23:04

Exercise 2 in Chapter 1 of Krylov, Mikhalev, and Tuganbaev's Endomorphism Rings of Abelian Groups asks to show that $\mathbb{F}_p \times \mathbb{F}_p$ is not the endomorphism ring of any abelian group. This is pretty clear: since $\mathbb{F}_p$ acts on the group $G$, as Kevin says $G$ must be an $\mathbb{F}_p$-vector space.

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Since p times the identity endomorphism on $\mathbb{Z}_p \times \mathbb{Z}_p$ is equal to zero, I don't see why the corresponding group $G$ would have to be cyclic. – Ricky Demer Jul 17 '10 at 7:34

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