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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.

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3 Answers 3

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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.

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

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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.

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