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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.

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).

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?

In other words, what do I need to impose on $A$ in order to make it a PID module? If $A$ is a $P$-module with $P$ a PID, then $P\subset R\subset Q(P)$ (the quotient field, am I right?), and so $R$ is a PID. Thank you.

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.

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.

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?

In other words, what do I need to impose on $A$ in order to make it a PID module? If $A$ is a $P$-module with $P$ a PID, then $P\subset R\subset Q(P)$ (the quotient field, am I right?), and so $R$ is a PID. Thank you.