0

Let $A$ be an augmented commutative $k$-algebra, where $k$ is a commutative ring. Let $\varepsilon:A\to k$ be the augmentation and $\eta:k\to A$ be the unit. Let $M$ be a right $A$-module.

Is it true that $\mathrm{Tor}_*^A(M_{\eta\varepsilon},k_\varepsilon)\cong M \otimes_k \mathrm{Tor}^A_*(k_\varepsilon,k_\varepsilon)$?

Here the subindexes indicate module structures, i.e. $M_{\eta\epsilon}$ is an $A$-module through $\eta\circ \epsilon$ and $k_\varepsilon$ is an $A$-module through $\varepsilon$.

user46225
  • 731
  • 4
  • 11
  • If you don't know if $M_{\eta}$ is $k$-flat, at least you have the converging spectral sequence $\mathrm{Tor}^k_p(\mathrm{Tor}^A_q(k_{\varepsilon},k_{\varepsilon}),M_{\eta}) \Rightarrow \mathrm{Tor}^A_{p+q}(M_{\eta\varepsilon},k_\varepsilon)$. – SlavaM Dec 01 '15 at 21:50
  • @Heinrich: could you please expand that into a full answer? What is that spectral sequence? The flatness hypothesis I'm ready to assume, though. – user46225 Dec 01 '15 at 22:31
  • Ok, I think it might be the one called "base change for Tor" on Weibel, theorem 5.6.6. I'll try to figure it out by myself from here, but if you want to expand your comment into an answer I'd be happy to upvote and accept it. – user46225 Dec 01 '15 at 22:39
  • Ok, I got it, it really is just that. Also, the $M_\eta$ is a red herring, might as well take any $k$-module. I didn't know this spectral sequence. Thanks a lot for the pointer. – user46225 Dec 01 '15 at 23:02

0 Answers0