There is an interesting result in Rotman's homological algebra book. Let $A$ be an $R$-module, $B$ be an $(R,S)$-bimodule and $C$ an $S$-module. (All rings commutative). Then Corollary 10.61 (in the 2009 version) states
If ${}_R B_S$ is flat on either side then $$\text{Tor}^S_n(A \otimes_R B,C) \simeq \text{Tor}^R_n(A,B \otimes_S C)$$
If particular if we let $B = S$ (assuming this is also an $R$-module but not necessarily $R$-flat) then $$\text{Tor}^S_n(A \otimes_R S,C) \simeq \text{Tor}^R_n(A,C)$$
Yet if we start with a projective resolution of $A$, $P^\bullet \to A$, $\text{Tor}^R_*(A,C)$ is the homology of complex $P^\bullet \otimes_R C$. But it doesn't appear automatic that $P^\bullet \otimes_R S \to A \otimes_R S$ will be a projective resolution unless $S$ is $R$-flat.
What am I missing?
Edit: Here is an outline argument that Rotman gives. If $A_R$ and ${}_RB_S$ satisfy $\text{Tor}^R_i(A,B\otimes_S,P) = 0$ for all $i \ge 0$ whenever ${}_S P$ is projective, then there is a spectral sequence
$$\text{Tor}_p^R(A,\text{Tor}_q^S(B,C))\Rightarrow \text{Tor}_n^S(A \otimes_RB,C)$$
Now if $B_S$ is flat and ${}_SP$ is projective (hence flat) then $B \otimes_S P$ is a flat $R$-module$^\ast$, then the hypothesis of the spectral sequence holds, and it collapses to the desired isomorphism.
($^\ast$Is this true?)
Best Answer
It should be part of the hypotheses that $S$ is a flat $R$-module. I doubt the theorem is true for general ring extensions.
I suspect that what Rotman means by "flat on either side" is that it is flat on both sides.