# Projective modules and ring homomorphisms.

Let $\varphi:R\rightarrow S$ be a ring homomorphism between commutative rings with unity. If $P$ is a projective $R$-module, is its extension $P\otimes_RS$ a projective $S$-module?

I tried shows that the functor $Hom_S(P\otimes_RS,\_)$ is exact, but I didn't know as find a homomorphism which the induced homomorphism carries it on a fixed homomorphism.

Yes, every projective module is a direct summand of a free module. If $P$ is projective, then $P\oplus Q=F$ for some module $P$ and some free module $F$. Tensoring with $S$ gives $(P\otimes_RS)\oplus(Q\otimes_R S)=F\otimes_R S$. As a direct sum of a number of copies of $R\otimes_R S$, $F\otimes_R S$ is free over $S$. As a direct summand of a free module, $P\otimes_R S$ is free.

• There are some isomorphisms of $R$-modules. Are they isomorphisms of $S$-modules also? Nov 26, 2017 at 18:28

Extension of scalars is left-adjoint to restriction of scalars. Restriction of scalars preserves epimorphisms. It follows from abstract nonsense that extension of scalars sends projective objects to projective objects.

I wanted to give another proof that I found nice.

Suppose $$\varphi:R\to S$$ is a homomorphism of commutative rings, and consider $$S$$ with the natural $$R$$-Module structure induced by $$\varphi$$.
If $$P$$ is $$R$$-Projective, then $$P\otimes_R S$$ is $$S$$-Projective.

Proof: The Hom-Tensor adjunction gives us a natural isomorphism of functors $$\operatorname{Hom}_S(P\otimes_R S,-) \cong \operatorname{Hom}_R(P,\operatorname{Hom}_S(S,-)).$$

Since $$P$$ is $$R$$-Projective, the functor $$\operatorname{Hom}_R(P,-)$$ is exact, and $$S$$ is always $$S$$-Projective, so $$\operatorname{Hom}_S(S,-)$$ is also exact, so the right-hand side is a composition of exact functors, and hence exact.

Finally, Natural isomorphism preserves exactness, so we see that $$\operatorname{Hom}_S(P\otimes_R S,-)$$, is exact, hence $$P\otimes_R S$$ is $$S$$-Projective.