Does every ring with unity arise as an endomorphism ring?

Solution 1:

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

Solution 2:

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.