Module structure on $R$-algebra $S$ by restriction of scalars and splitting of the algebra map

Question

Let $R, S$ be commutative Noetherian rings. Let $f: R \to S$ be a ring homomorphism. Consider the $R$-module $f_* S$ whose underlying abelian group is $S$ itself but the $R$-module structure is given by $r\cdot s:= f(r)s, \forall r \in R, s \in f_*S$. With this module structure, note that $f: R\to f_*S$ is an $R$-linear map.

Now assume $f_*S$ is a finitely generated $R$-module. If $R$ is a direct summand of $f_*S$ as $R$-modules, then does the $R$-linear map $f: R\to f_*S$ split as a morphism of $R$-modules ?

Answer 1

Yes.

Note that $f_\ast S$ can be regarded as $R\otimes_RS$, with $f:R\to R\otimes_RS$ given by $f(r)=r\otimes1$.

The map $g:S\to S\otimes_RS$ given by $g(s)=s\otimes1$ is a split $R$-module monomorphism, with splitting map the multiplication map $m:S\otimes_RS\to S$ given by $m(s\otimes s')=ss'$.

Suppose $S=R\oplus X$ as $R$-modules. So $g$ is the direct sum of the map $f:R\to R\otimes_R S$ and a map $h:X\to X\otimes_RS$ given by $h(x)=x\otimes1$.

Then $\pi\circ m:R\otimes_RS\to R$, where $\pi:R\oplus X\to R$ is projection onto the direct factor, is a splitting map for $f$.