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I don't understand a step of a proof in Weibel's book. Proposition 3.2.9 in Weibel's homological algebra book states that

Assume $T$ is a flat $R$-algebra. Then for all $T$-modules $C$ and $R$-modules $A$ it holds that \begin{align*} Tor^R_n(A,C) = Tor_n^T(A\otimes_R T,C)\,. \end{align*}

The proof goes as follows. Take a projective resolution $P_*$ of $A$. Since $T$ is flat $P_*\otimes_R T$ is a resolution of $A \otimes_R T$. Next Weibel states that each $P_n \otimes_R T$ is a projective $T$-module. I don't see why this is the case. Question: Why are the $P_n\otimes_RT$ projective?

My thoughts: I know that left adjoints preserve projective objects when their right adjoint is exact. But to use this I would need that $Hom(T,-)$ is exact, i.e. that $T$ is projective, and flat modules are not projective in general. I think the statement can be saved by using that $Tor$ can be computed by using flat resolutions. If $f$ is monic, then so is $(P_n \otimes_R T)\otimes_T f$ since tensor product is associative and $P_n$ is flat.

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You need to pay attention to the category you are working in. $T$ is certainly a projective $T$-module, and $\textrm{Hom}_T (T, -)$ is exact as a functor from $T$-modules to $R$-modules, so ${-} \otimes_R T$ sends projective $R$-modules to projective $T$-modules.

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  • $\begingroup$ Oh, you are right, thank you. I stumbled upon many of your answers and questions recently and found them quite helpful. I am learning the same things at the moment which you learned years ago. In a view years I'll be a grown up mathematician too. (: $\endgroup$
    – Nico
    Commented Jun 8, 2021 at 11:33

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