TITLE: Is every ring a homomorphic image of some abelian group's endomorphism ring?
QUESTION [9 upvotes]: Question:

Is every ring a homomorphic image of some abelian group's  endomorphism ring?

I'm a beginner in ring theory. I don't know if this is true and google  didn't really help me out here.

REPLY [6 votes]: The answer is "yes".
Every ring is a homomorphic image of the free ring $\mathbb{Z}\langle X\rangle$ generated by some sufficiently large set $X$. So it suffices to prove that $\mathbb{Z}\langle X\rangle$ is the endomorphism ring of some abelian group.
There may be a much more elementary way to show this, but it follows from the main theorem of
Dugas, Manfred; Göbel, Rüdiger, Every cotorsion-free algebra is an endomorphism algebra, Math. Z. 181, 451-470 (1982). ZBL0501.16031,
which (specialized from the more general context of modules for a Dedekind domain $R$ to the case $R=\mathbb{Z}$) states that every ring whose additive group is a cotorsion-free abelian group is the endomorphism ring of some abelian group.
Since the additive group of $\mathbb{Z}\langle X\rangle$ is a free abelian group, and therefore cotorsion-free, it follows that $\mathbb{Z}\langle X\rangle$ is the endomorphism ring of some abelian group.