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I don't believe that every ring with a $1$ is the endomorphism ring of an abelian group but I currently don't see how to produce a counterexample.

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Related question on MathOverflow: mathoverflow.net/questions/27971/… –  Jonas Meyer Dec 8 '12 at 22:48
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Every ring $R$ is the ring of all endomorphisms of the $R$-module $R$. I'm not sure if that's along the lines you're thinking. –  Hurkyl Dec 8 '12 at 23:00
    
What might be $G$ with $\operatorname{End}(G)\cong\mathbb Z[X]$? –  Hagen von Eitzen Dec 8 '12 at 23:04
    
@Hagen: I can try to engineer some model where the axiom of choice fails... :-) –  Asaf Karagila Dec 8 '12 at 23:08
    
Why did the $\Bbb Z_2\times \Bbb Z/2$ approach vanish? The argument was not perfectly sound, but I think the idea was the right one: large abelian groups yield large endomorphism rings so some small rings which are not attained by small groups are left out. –  Dominik Dec 9 '12 at 1:50

2 Answers 2

up vote 7 down vote accepted

Proposition. Let $A = \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Then $A$ is not the endomorphism ring of an abelian group.

Proof: Suppose $A$ is the endomorphism ring of an abelian group $G$. Let $e = (1, 1)$ be the unity of $A$. Then $2e = 0$. Let $x \in G$. Then $2x = 2(ex) = (2e)x = 0$. Hence $G$ can be regarded as a vector space over a field $k = \mathbb{Z}/2\mathbb{Z}$. Then $A = \operatorname{End}_k(G)$. Suppose $n = \dim_k G$. Then $|A| = 2^{n^2}$. This is impossible. QED

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A similar proof can be applied to $(\mathbb{Z}/p\mathbb{Z})\times (\mathbb{Z}/p\mathbb{Z})$, where $p$ is a prime number. –  Makoto Kato Dec 9 '12 at 2:17
    
Or you could use $F_{p^2}$. –  Dominik Dec 9 '12 at 11:03
    
@Dominik I don't think I understand. Could you explain it? –  Makoto Kato Dec 9 '12 at 13:12
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Since the ring $\Bbb F_{p^2}$ has $p=0$ we obtain $px=0$ for every $x \in G$. So $G$ can be regarded as a $\Bbb Z / p\Bbb Z$ vector space, yielding $|A|=p^{dim_{F_p}(G)^2}>p^2$. –  Dominik Dec 9 '12 at 13:20
    
@Dominik Right. Thanks. –  Makoto Kato Dec 9 '12 at 13:26

I came up with an example of an infinite ring which is not the endomorphism ring of an abelian group.

Proposition Let $K$ be a division ring of characteristic $p > 0$. Suppose $|K| > p \ (K$ may be an infinite division ring). Then $K$ is not the endomorphism ring of an abelian group.

Proof: Suppose $K$ is the endomorphism ring of an abelian group $G$. Since $p = 0$ in $K$, $G$ can be regarded as a vector space over $F = \mathbb{Z}/p\mathbb{Z}$. Since $|K| \neq p$, $\dim_F G > 1$. Hence $G$ has a non-zero proper $F$-subspace $H$. Hence there exists a non-zero proper $F$-subspace $H'$ such that $G = H \oplus H'$. Let $f$ be the projection map $G \rightarrow H$ induced by the decomposition $G = H \oplus H'$. Since $f$ can be regarded as an element of $K = End(G)$ and $f^2 = f$, $f = 1$. This is a contradiction. QED

EDIT I found that the above idea can be applied to a division ring of characteristic $0$ except $\mathbb{Q}$.

Proposition 2 Let $K$ be a division ring. Let $F$ be the prime subfield of $K$. Suppose $(K \colon F) > 1$. Then $K$ is not the endomorphism ring of an abelian group.

Proof: Suppose $K$ is the endomorphism ring of an abelian group $G$. Then $G$ can be regarded as as a vector space over $K$. Hence it can be regarded as as a vector space over $F$. Since $(K \colon F) > 1$, $\dim_F G > 1$. Hence $G$ has a non-zero proper $F$-subspace $H$. Hence there exists a non-zero proper $F$-subspace $H'$ such that $G = H \oplus H'$. Let $f$ be the projection map $G \rightarrow H$ induced by the decomposition $G = H \oplus H'$. Since $f$ can be regarded as an element of $K = End(G)$ and $f^2 = f$, $f = 1$. This is a contradiction. QED

EDIT 2 I found a large class of rings which are not the endomorphism rings of abelian groups.

Let $A$ be a ring. An element $e$ of $A$ is called an idempotent if $e^2 = e$. An idempotent which is neither $0$ nor $1$ is called a non-trivial idempotent. If $e$ is an idempotent, $f = 1 - e$ is also an idempotent.

Lemma 1 An integral domain has no non-trivial idempotents.

Proof: Let $A$ be an integral domain. Let $e$ be an idempotent of $A$. Then $e(1 - e) = 0$. Hence $e = 0$ or $1$. QED

Lemma 2 A local ring has no non-trivial idempotents.

Proof: Let $A$ be a local ring. Let $\mathfrak{m}$ be the maximal ideal of $A$. Let $e$ be an idempotent. If $e$ is an invertible element, $e = 1$. Suppose $e$ is not invertible. Then $e \in \mathfrak{m}$. Hence $1 - e$ is invertible. Since $1 - e$ is an idempotent, $1 - e = 1$. Hence $e = 0$. QED

Proposition 3 (generalization of propositions 1, 2) Let $A$ be an algebra over a field $K$. Let $F$ be the prime subfield of $K$. Suppose $\dim_F A > 1$. Suppose $A$ has no non-trivial idempotents (for example, $A$ is a division ring or an integral domain or a local ring). Then $A$ is not the endomorphism ring of an abelian group.

Proof: The same as the proof of proposition 2.

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Dear Makoto, I find it great that you spend so much time answering questions on this site, but I think perhaps more people would read them (and upvote them) if they all weren't so long and formal. (for example, most people reading this thread knows what an idempotent is, and if not, they can consult Wikipedia). –  Fredrik Meyer Dec 10 '12 at 17:59
    
@FredrikMeyer One is too often tempted to write a sketchy(hence incomplete and hard to understand) proof because it looks cool and makes you look smarter and most of all, saves your time. I disagree with that attitude. –  Makoto Kato Dec 10 '12 at 18:34

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