Let $R$ be a ring (assumed associative and unital) whose additive group is a finitely generated abelian group. As a reduction step in a paper I'm working on, we need to know that $R$ is a quotient of another ring $S$ whose additive group is a finite rank free abelian group. We believe we have a proof, but it is rather involved. Is this something that has appeared already or at least has a reasonably short proof? (The current proof reduces to showing it for finite rings and then proves it by using the basic ring associated to $R$.)


When $R$ is commutative, this has a quick proof: $R$ is generated by $r\_1,\dots,r\_n$ which are integral over ${\bf Z}$, and hence $R$ is a quotient of ${\bf Z}[x\_1,\dots,x\_n]/(f\_1(x\_1),\dots,f\_n(x\_n))$ where each $f\_i$ is a monic univariate polynomial. 


So I'm wondering if there is something analogous in the general case.



Quotients of rings with finite free additive group
For a finite ring $R$ you can write it as a quotient of the monoid ring $\mathbb ZR$ (with respect to the multiplicative structure) which has a finitely generated free additive group. 


