## 15.60 Spectral sequences for Tor

In this section we collect various spectral sequences that come up when considering the Tor functors.

Example 15.60.1. Let $R$ be a ring. Let $K_\bullet$ be a chain complex of $R$-modules with $K_ n = 0$ for $n \ll 0$. Let $M$ be an $R$-module. Choose a resolution $P_\bullet \to M$ of $M$ by free $R$-modules. We obtain a double chain complex $K_\bullet \otimes _ R P_\bullet$. Applying the material in Homology, Section 12.25 (especially Homology, Lemma 12.25.3) translated into the language of chain complexes we find two spectral sequences converging to $H_*(K_\bullet \otimes _ R^\mathbf {L} M)$. Namely, on the one hand a spectral sequence with $E_2$-page

$(E_2)_{i, j} = \text{Tor}^ R_ j(H_ i(K_\bullet ), M) \Rightarrow H_{i + j}(K_\bullet \otimes ^{\mathbf{L}}_ R M)$

and differential $d_2$ given by maps $\text{Tor}^ R_ j(H_ i(K_\bullet ), M) \to \text{Tor}^ R_{j - 2}(H_{i + 1}(K_\bullet ), M)$. Another spectral sequence with $E_1$-page

$(E_1)_{i, j} = \text{Tor}^ R_ j(K_ i, M) \Rightarrow H_{i + j}(K_\bullet \otimes ^{\mathbf{L}}_ R M)$

with differential $d_1$ given by maps $\text{Tor}^ R_ j(K_ i, M) \to \text{Tor}^ R_ j(K_{i - 1}, M)$ induced by $K_ i \to K_{i - 1}$.

Example 15.60.2. Let $R \to S$ be a ring map. Let $M$ be an $R$-module and let $N$ be an $S$-module. Then there is a spectral sequence

$\text{Tor}^ S_ n(\text{Tor}^ R_ m(M, S), N) \Rightarrow \text{Tor}^ R_{n + m}(M, N).$

To construct it choose a $R$-free resolution $P_\bullet$ of $M$. Then we have

$M \otimes _ R^{\mathbf{L}} N = P^\bullet \otimes _ R N = (P^\bullet \otimes _ R S) \otimes _ S N$

and then apply the first spectral sequence of Example 15.60.1.

Example 15.60.3. Consider a commutative diagram

$\xymatrix{ B \ar[r] & B' = B \otimes _ A A' \\ A \ar[r] \ar[u] & A' \ar[u] }$

and $B$-modules $M, N$. Set $M' = M \otimes _ A A' = M \otimes _ B B'$ and $N' = N \otimes _ A A' = N \otimes _ B B'$. Assume that $A \to B$ is flat and that $M$ and $N$ are $A$-flat. Then there is a spectral sequence

$\text{Tor}^ A_ i(\text{Tor}_ j^ B(M, N), A') \Rightarrow \text{Tor}^{B'}_{i + j}(M', N')$

The reason is as follows. Choose free resolution $F_\bullet \to M$ as a $B$-module. As $B$ and $M$ are $A$-flat we see that $F_\bullet \otimes _ A A'$ is a free $B'$-resolution of $M'$. Hence we see that the groups $\text{Tor}^{B'}_ n(M', N')$ are computed by the complex

$(F_\bullet \otimes _ A A') \otimes _{B'} N' = (F_\bullet \otimes _ B N) \otimes _ A A' = (F_\bullet \otimes _ B N) \otimes ^{\mathbf{L}}_ A A'$

the last equality because $F_\bullet \otimes _ B N$ is a complex of flat $A$-modules as $N$ is flat over $A$. Hence we obtain the spectral sequence by applying the spectral sequence of Example 15.60.1.

Example 15.60.4. Let $K^\bullet , L^\bullet$ be objects of $D^{-}(R)$. Then there are spectral sequences

$E_2^{p, q} = H^ p(K^\bullet \otimes _ R^{\mathbf{L}} H^ q(L^\bullet )) \Rightarrow H^{p + q}(K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet )$

with $d_2^{p, q} : E_2^{p, q} \to E_2^{p + 2, q - 1}$ and

$H^ q(H^ p(K^\bullet ) \otimes _ R^{\mathbf{L}} L^\bullet ) \Rightarrow H^{p + q}(K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet )$

After replacing $K^\bullet$ and $L^\bullet$ by bounded above complexes of projectives, these spectral sequences are simply the two spectral sequences for computing the cohomology of $\text{Tot}(K^\bullet \otimes L^\bullet )$ discussed in Homology, Section 12.25.

Comment #5790 by Brad Dirks on

Is there a typo in the indices for Example 061Z? Homological spectral sequences have $d_r$ with bidegree $(-r,r-1)$, so $d_2$ should go from $E_{p,q}^2$ to $E_{p-2,q+1}^2$. This would be a map $Tor_q( H_p(K_\bullet), M) \rightarrow Tor_{q+1}(H_{p-2}(K_\bullet),M)$. It seems that the $E_1$ page in the same example does have the correct indices.

Comment #5805 by on

I think you are right about the bidegree of $d_r$ being $(-r , r - 1)$ but I think you swapped the roles of $i$ and $j$. So I think the differential for the first spectral squence of the example on the $E_2$-page is given by maps $Tor_j(H_i, M) \to Tor_{j - 2}(H_{i + 1}, M)$ which is what it says.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).