Does the restriction of scalars functor preserve the quotient module construction?

Question

I have two rings $R,\,S$ (not necessarily commutative - in the case I have in mind, just $R$ is commutative) such that $i:R\subset S$. Clearly, if we have a module $M$ over $S$, then we can restrict this to a module over $S$ (call it $i^\ast(M)$) using the normal rules of restriction of scalars, i.e. $rs=i(r)s$. I was wondering if this sort of restriction ever preserves the quotient; that is, if we have an $S$-module $M/N$, then does the $R$-module satisfy $i^\ast(M/N)\cong i^\ast(M)/i^\ast(N)$?

Answer 1

If $$0\to A\to B\to C\to 0$$ is an exact sequence of $S$-modules, then $$0\to i^\ast(A)\to i^\ast(B)\to i^\ast(C)\to 0$$ is an exact sequences of $R$-modules, because it has the same underlying sets and the same functions.