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.

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
