Suppose that $R$ is a ring, $I \subset R$ is an ideal, and $f : A \to B$ is a homomorphism of $R$-modules. Is it true that there is an induced homomorphism $f/I : A/IA \to B/IB$ which is a map of $R/I$-modules?
I realize this should be easy to check (or dis-check), but I'm having a hard time figuring out what exactly needs to be verified. I know $f(x+iy) = f(x) + i f(y)$ which is the same thing in $B/IB$, so I get a map of $R$-modules at least. What does it mean to be a map of $R/I$-modules?
Hint.
$$f : A \to B$$ now tensor with $R/I$: $$f\otimes_R1_{R/I}:A\otimes_R R/I \to B\otimes_R R/I $$ and use $A\otimes_R R/I \equiv A/IA$.
(this even shows that if $f$ is surjective then $f/I$ is surjective too)