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

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

I ask because I've never liked to identify rings as being subrings of endomorphism rings. A subring is basically a ring within another ring, so if you answer "what is natural about the ring axioms (and by extension, rings as a whole)" with "because they are (essentially) subrings of an endomorphism ring", then to me it feels like saying "rings are naturals because they are rings"... well yes, but why those axioms?

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

• Endomorphism of whatsits? Every ring can be realized as a subring of the endomorphism ring of its underlying abelian group... May 11, 2022 at 20:53
• Working on a question whose truth/falsity you are not sure of is like the purest form of doing math there is. I'm not going to do it, but I was tempted to try to close this question with a reason of "This will build character". :-) May 11, 2022 at 21:36
• @rschwieb Wouldn't the identity need to be $R=\operatorname{End}(R_{\Bbb Z})$ (see Arturo's comment in that regard) in order for your remark to answer the question? May 12, 2022 at 6:17
• @PineappleFish You should not ask the question only in the title. It should be clear in the post too. It is fine to be in both the title and post. You should also consider removing some of the chattiness from the body... it's only contributing fluff to your post. It's enough to say you're a beginner to ring theory, that you don't know if this is true, and that you haven't been able to find an answer in google. Ideally also you'd consider some examples and why you ruled them out, but I understand that's hard when you're a beginner. May 12, 2022 at 14:37
• This is an excellent question that appears to be pretty difficult. I would guess the answer is no, based on the vague intuition that endomorphism rings tend to be "close to simple", but non-finitely generated abelian groups are weird and can do all sorts of weird things. May 13, 2022 at 23:29

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.
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.