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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?)

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2 Answers 2

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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.

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  • $\begingroup$ Thanks. I've always thought this was the case. The proof he gives is definitely only on one side. I've sketched it above, along with the part that I don't quite see. $\endgroup$
    – Drew
    Feb 28, 2013 at 0:24
  • $\begingroup$ The *'d comment in your edit is certainly true when $_RB_S$ is flat as an $R$-module. It is not in general true when $_RB_S$ is not flat as an $R$-module. $\endgroup$
    – Jim
    Feb 28, 2013 at 0:27
  • $\begingroup$ Right. Which is exactly what Rotman (seems to be) assuming $\endgroup$
    – Drew
    Feb 28, 2013 at 0:34
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This is true in general (assuming, as I'm sure was meant, that $B$ is flat "on either side" means on both sides), and I don't think you need the rings to be commutative. If $P_* \to A$ is a free resolution, then $P_* \otimes_R B$ has homology groups $Tor^R(A,B)$, and so it's also exact above degree zero. It's then a resolution too -- not by free or projective objects, but (since $P_*$ was free) by direct sums of copies of $B$. This makes $P_* \otimes_R B$ into a resolution of $A \otimes_R B$ by flat right $S$-modules, and that's good enough to compute Tor.

This makes the homology of the complex $(P_* \otimes_R B) \otimes_S C$ into something that computes $Tor_S(A \otimes_R B, C)$.

On the other hand, it is isomorphic to the complex $P_* \otimes_R (B \otimes_S C)$, which more directly is something that computes $Tor_R(A,B \otimes_S C)$.

(This feels a little ad-hoc. One perspective on Tor is that you're allowed to take a resolution of one side, or the other side, or of both sides and tensor together the resolutions; this is how you usually show that Tor over a commutative ring is symmetric. For this problem, that means that you can take resolutions of $A$ and $C$ simultaneously, which from a certain perspective might clean this argument up.)

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  • $\begingroup$ Thanks Tyler. I agree with what you are saying. (I emailed Rotman and it appears it is just an error in the book) $\endgroup$
    – Drew
    May 14, 2013 at 3:24

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