# When does a ring homomorphism yield module structure?

Assume all rings have 1, preserved by homomorphisms, and that all modules are unitary.

Given a ring homomorphism $$\phi:R\to S$$, any $$S$$-module A can be given an $$R$$-module structure by the action $$ra\mapsto\phi(r)a$$.

Given again the ring homomorphism $$\phi$$, what other conditions are needed to utilize $$\phi$$ for giving any $$R$$-module an $$S$$-module structure?

• You may want to edit your question's title, because as is it seems like the content of the post is in itself the answer you are looking for (i.e. it would be useful to state that you look for structure on 'the other direction'). – Guido A. Oct 7 '18 at 7:05

## 1 Answer

For the ring homomorphism $$\phi: R \to S$$ you have described the functor of “restriction along $$\phi$$”, which takes an $$S$$-module and views it as an $$R$$ module by using $$\phi$$. There is a functor in the opposite direction, called “induction along $$\phi$$”, which takes any $$R$$-module and produces an $$S$$-module.

Induction is slightly trickier to explain. We can view $$S$$ as an $$(S, R)$$ bimodule, via the actions $$s \cdot x = sx$$ and $$x \cdot r= x \phi(r)$$. Then, if $$V$$ is a left $$R$$-module, the tensor product $$S \otimes_R V$$ is the “induction of $$V$$ along $$\phi$$”.