Abelian groups such that $A \cong \mathrm{End}(A)$ and "complete rings"

Question

Motivation: for any ring $R$ there is the natural monomorphism $\mathrm{in} \colon R \to \mathrm{End}(R_{add}): r \mapsto (x \mapsto rx)$, where $R_{add}$ is an additive abelian group ( rings are assumed to be associative with identity, but not necessarily commutative). So a ring is exactly an abelian group with a distinguished subgring of its endomorphisms (and a fixed bijection between elements and distinguished endomorphisms). Some rings are "complete" in the sense that they "contain" all endomorphisms of the underlying abelian group. For example, $\mathrm{in}$ is an isomorphism for $R = \mathbb{Z}, \mathbb{Q}, \mathbb{Z}_n$.

1. What is known about the classification of abelian groups $A$ such that there is an isomorphism between the abelian groups $\mathrm{End}(A)$ and $A$?

Each such isomorphism gives some "complete ring" structure on $A$.

2. What is known about the uniqueness (up to isomorphism) of the "complete ring" structure on an abelian group?

I'm interested in the answers to these questions, with any additional assumptions that seem natural to you. I am especially interested in the answers for commutative rings.

Answer 1

The rings, you call ``complete'' are known as $E$-rings (as Ulrich Pennig mentioned in the comments).

Some comments on your questions

1. There are too many results on the $E$-rings to list them here and I'd rather direct you to the book by Göbel and Trlifaj Approximations and endomorphism algebras of modules. However to give you some sense that we have no hope to obtain any reasonable classification - every Abelian cotorsion-free group embeds in an $E$-ring. Formally a group $A$ is cotorsion-free if there are only null homomorphisms $\mathbb{Z}^\wedge_p\to A$. This is equivalent to the claim that if another Abelian group $C$ admits a compact topology then every homomorphism $C\to A$ is null. To make long story short - in the absence of compactness you may add to $A$ new elements to get more endomorphisms and then add yet more elements to kill the unwanted endomorphisms.
2. Given an Abelian group $A$ the $E$-ring structure on $A$ is unique up to the choice ot the identity element. Every invertible element of an $E$-ring $A$ can be chosen as the identity element for another ring structure on $A$.

Answer 2

In the finitely generated case, it's relatively easy to see from the structure theorem that the only finitely generated abelian groups $A$ with an isomorphism $A \cong \text{End}(A)$ are the cyclic ones.

Here is a sketch of the proof of this claim : If $B = \mathbb{Z}/n_1\mathbb{Z} \times \cdots \times \mathbb{Z}/n_k\mathbb{Z}$ (where $n_1 \mid n_2 \mid \cdots \mid n_k$ and $n_i > 1$) is the torsion part of $A$, then the torsion part of $\text{End}(A)$ contains $\text{End}(B)$. Moreover $B$ embeds in $\text{End}(B)$ via the natural ring structure on $B$, but if $k>1$ then $\text{End}(B)$ contains more elements, for example the endomorphism sending the generator of $\mathbb{Z}/n_2\mathbb{Z}$ to the generator of $\mathbb{Z}/n_1\mathbb{Z}$ and the generators of all the other $\mathbb{Z}/n_i\mathbb{Z}$ to zero. Hence if $k > 1$ the torsion part of $\text{End}(A)$ has larger size than the torsion part of $A$, so $\text{End}(A) \not\cong A$.

Similarly if $A=\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}^r$ for $n > 1$, $r>0$, then it's also easy to see that the torsion part of $\text{End}(A)$ has size larger than $n$. Finally, if $A=\mathbb{Z}^r$ for $r>2$, then $\text{End}(A) \cong M_r(\mathbb{Z}) \cong \mathbb{Z}^{r^2} \not\cong A$.

That leaves only the cases $A=\mathbb{Z}/n\mathbb{Z}$ or $A=\mathbb{Z}$, i.e. $A$ must be cyclic. In that case, question 2 is also easy to answer, as all ring structures on $A$ are the same up to an automorphism of $A$ (actually, this is true whenever $\text{End}(A)$ is a commutative ring).