# 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).

• I am sure you know this, but someone has to say it: finite generation is sufficiant. Apr 22, 2016 at 13:08
• Yes, thank you for the comment. For $A$ f.g we know how $R$ looks like. Apr 22, 2016 at 13:14

Two classes of torsion-free abelian groups having the desired property are

1. free abelian groups
2. 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 $R$.

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.

• Thanks a lot for the answer and for the reference. I will need to digest it. Sep 8, 2016 at 14:02