The center of a(n endomorphism) ring is a PID Let $A$ be a torsion-free Abelian group, and let $\mbox{End}(A)$ be its endomorphism ring. Denote by $R\doteq\mathbf{Z}(\mbox{End}(A))$ the center of $\mbox{End}(A)$. What would be necessary and/or sufficient conditions on $A$ for $R$ to be a PID? Thank you.
Edit Following  user89334's comment, indeed, this is true if $A$ is a finitely generated group. The core of the question is the infinitely generated case (no restrictions on the cardinality of the generating set).
 A: Two classes of torsion-free abelian groups having the desired property are 


*

*free abelian groups 

*torsion-free divisible groups (here I use the axiom of choice) 


By noting that a torsion-free divisible group is a $\mathbb{Q}$-vector space and that a $\mathbb{Z}$-linear endomorphism of such a group is  $\mathbb{Q}$-linear, the two examples follows from the easy to prove 

For a ring $R$ with identity and a free $R$-module $F$, the center of a $End_R(F)$ is isomorphic (as ring) to the center of $R$. 

A: Here are two results from the literature, found in Endomorphism Rings of Abelian Groups, p. 269. For these to be relevant, we need $\mathrm{End}(A)$ to be commutative.
Some terminology. A group is $A$-free if it is a direct sum of copies of $A$, and a group is $A$-projective if it is a direct summand of an $A$-free group.
We have:

If $\mathrm{End}(A)$ is a PID, then every $A$-projective
  group is $A$-free.

Next, call $A$ self-small if for all indexing sets $I$, the image of every homomorphism $A\rightarrow\bigoplus_{i\in I} A$ lies in a finite sum. There is a lot of discussion regarding this condition in the aforementioned reference, pp. 254-263.
The result is:

Suppose $A$ is self-small. Then $\mathrm{End}(A)$ is a PID if and only if every projective right $\mathrm{End}(A)$-module is free.

