This question came up in this thread: Proving that tensoring a projective module with a flat module gives a projective module?

Assume $\left\{P_i\right\}$ is a projective resolution of an $R$-module $M$. Why is it that, if $R\to S$ is a ring homomorphism that makes $S$ into a flat $R$-module, that $\left\{P_i\otimes_R S\right\}$ is a projective resolution for $M\otimes_R S$?

  • 2
    $\begingroup$ I think it's because tensoring with a flat module preserves exact sequences. $\endgroup$ – Tunococ Jul 8 '14 at 21:49
  • $\begingroup$ That's true, but why do the modules remain projective? $\endgroup$ – Nishant Jul 8 '14 at 21:53
  • $\begingroup$ Oh, I thought that you were allowed to used the fact from the linked thread. I may have misunderstood. $\endgroup$ – Tunococ Jul 8 '14 at 21:57
  • $\begingroup$ I am confused does not the linked thread conclude that tensor of projective and flat may not be projective. Is the situation here somehow different ? $\endgroup$ – Rene Schipperus Jul 8 '14 at 22:06
  • $\begingroup$ @Nishant $P \otimes_R S$ is a projective $S$-module if $P$ is a projective $R$-module. $\endgroup$ – Zhen Lin Jul 8 '14 at 22:08

This is true with added generality, that is in the non commutative world. So, assume $\phi\colon R\to S$ is a ring homomorphism between not necessarily commutative rings. If $P_{R}$ is a projective right $R-$module, then, if $P_{R}$ is a direct summand of the free right module $R^{(X)}$ for some set $X$, we get that

$$ P_{R}\otimes_{R} S $$

is a direct summand of $R^{(X)}\otimes_{R} S\simeq S^{(X)}$, thus it is a projective right $S-$module. It follows that tensoring with $S$ sends projective resolutions of a right $R-$module $M_{R}$ into $S-$projective resolutions of $M_{R}\otimes_{R} S$. In particular, if $M_{R}$ is projective, so is $M_{R}\otimes_{R} S$ (as an $S$-module).

I would like to mention that this allows to prove what follows. Under the above hypothesis of flatness of $S$, if $C$ is a left $S-$module, then, for each $n\in \mathbb{N}$, we have

$$ Tor_{n}^{R}(M_{R},\ C)\simeq Tor_{n}^{S}(M_{R}\otimes_{R} S,\ C). $$ Indeed, the above argument says that, if $P_{\bullet}\to M\to 0$ is a projective resolution of $R$, then $$Tor_{n}^{S}(M_{R}\otimes_{R} S,\ C)\simeq H_{n}((P_{\bullet}\otimes_{R}S)\otimes_{S} C).$$ Since, for each (projective) $R-$module $P_{R}$, there is a natural isomorphism $$(P_{R}\otimes_{R}S)\otimes_{S} C\simeq P_{R}\otimes_{R}(S\otimes_{S} C)\simeq P_{R}\otimes_{R} C,$$ we get $$H_{n}((P_{\bullet}\otimes_{R}S)\otimes_{S} C)\simeq H_{n}(P_{\bullet}\otimes_{R} C),$$ which allows us to conclude.

As a corollary, we get that, if $R, S$ are commutative (with $S$ flat as an $R-$module) and $M,N$ are $R-$modules, then

$$Tor_{n}^{R}(M,\ N)\otimes_{R}S\simeq Tor_{n}^{S}(M\otimes_{R}S,\ N\otimes_{R}S).$$


Tunococ's comment provides half the answer: Flatness of $S$ implies that $\{P_i\otimes_RS\}$ is a resolution of $M\otimes_RS$. The other half is to observe that this resolution consists of projective $S$-modules. Indeed, each $P_i$ is a projective $R$-module, hence a summand of a free $R$-module. That direct sum decomposition is preserved by tensoring with $S$ (indeed, by any additive functor), and tensoring a free $R$-module with $S$ produces a free $S$-module.


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