For abelian groups, the existence of left inverse or right inverse of a homomorphism can be characterized by looking at whether the image or kernel splits the group. Is there an analogous characterization for ring homomorphisms? Specifically, if $f:R\rightarrow S$ be a ring homomorphism, when does there exist a ring homomorphism $g:S\rightarrow R$ such that $f\circ g=\mathrm{Id}_S$ or $g\circ f=\mathrm{Id}_R$ ?

Edit: Let's restrict to commutative rings for now.


I don't think there's anything nice to say in general. This is a fairly delicate problem.

Consider just the case of left inverses. Here is the sort of thing that can go wrong: take $f : k \to k[x]/f(x)$. Then a left inverse exists iff there exists $r \in k$ such that $f(r) = 0$, and e.g. if $k$ is a non-algebraically closed field this won't exist most of the time. More generally we can take maps like $f : \mathbb{Z} \to \mathbb{Z}[x_1, ..., x_n]/(f_1, f_2, ..., f_m)$ and we see that the problem of determining whether there exists a left inverse is at least as hard as the problem of determining whether a system of Diophantine equations has an integer solution, which is known to be undecidable in a particular sense by Matiyasevich's theorem. These examples show in particular that it doesn't suffice to assume that $f$ is a monomorphism or even that it makes $S$ a finite rank free $R$-module.

(Thinking of this question as a question about commutative rings, then as a question about affine schemes, note that the analogous question for spaces includes as a special case the question of whether a bundle admits a section, which is a pretty nontrivial question, e.g. it admits in turn as a special case the question of whether a manifold is parallelizable. I think one can rephrase this entirely as a question about C*-algebras if the manifold is compact, and already it's a nontrivial question to characterize the parallelizable spheres.)

Edit: I wrote a blog post exploring the topological analogue in more detail.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.