The Stacks project

15.62 Spectral sequences for Tor

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

Example 15.62.1. 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 15.62.2. 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 15.62.1.

Example 15.62.3. 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 15.62.1.

Example 15.62.4. Let $K^\bullet , L^\bullet $ be objects of $D^{-}(R)$. Then there is a spectral sequence with

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

and another spectral sequence with

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

Both spectral sequences have $d_2^{p, q} : E_2^{p, q} \to E_2^{p + 2, q - 1}$. 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.


Comments (2)

Comment #5790 by Brad Dirks on

Is there a typo in the indices for Example 061Z? Homological spectral sequences have with bidegree , so should go from to . This would be a map . It seems that the page in the same example does have the correct indices.

Comment #5805 by on

I think you are right about the bidegree of being but I think you swapped the roles of and . So I think the differential for the first spectral squence of the example on the -page is given by maps which is what it says.


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