Does every ring with unity arise as an endomorphism ring? 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.
 A: 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.
A: 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
