The Stacks project

Proof. As $K^\bullet $ is a bounded above complex of flat modules we see that $K^\bullet \otimes _ R M = K^\bullet \otimes _ R^{\mathbf{L}} M$. Hence for every $R$-module $M$ the sequence

is exact in the middle. Since $K^{a - 2} \to K^{a - 1} \to K^ a \to \mathop{\mathrm{Coker}}(d_ K^{a - 1}) \to 0$ is a flat resolution this implies that $\text{Tor}_1^ R(\mathop{\mathrm{Coker}}(d_ K^{a - 1}), M) = 0$ for all $R$-modules $M$. This means that $\mathop{\mathrm{Coker}}(d_ K^{a - 1})$ is flat, see Algebra, Lemma 10.75.8. $\square$

Proof. If (2) holds, then we may compute $K^\bullet \otimes _ R^\mathbf {L} M = E^\bullet \otimes _ R M$ and it is clear that (1) holds. Assume that (1) holds. We may replace $K^\bullet $ by a projective resolution with $K^ i = 0$ for $i > b$. See Derived Categories, Lemma 13.19.3. Set $E^\bullet = \tau _{\geq a}K^\bullet $. Everything is clear except that $E^ a$ is flat which follows immediately from Lemma 15.66.2 and the definitions. $\square$

Proof. The implication (2) $\Rightarrow $ (1) is immediate. Assume (1) holds. First we choose a K-flat complex $K^\bullet $ with flat terms representing $K$, see Lemma 15.59.10. For any $R$-module $M$ the cohomology of

computes $H^ n(K \otimes _ R^\mathbf {L} M)$. This is always zero for $n < a$. Hence if we apply Lemma 15.66.2 to the complex $\ldots \to K^{a - 1} \to K^ a \to K^{a + 1}$ we conclude that $N = \mathop{\mathrm{Coker}}(K^{a - 1} \to K^ a)$ is a flat $R$-module. We set

The kernel $L^\bullet $ of $K^\bullet \to E^\bullet $ is the complex

where $I \subset K^ a$ is the image of $K^{a - 1} \to K^ a$. Since we have the short exact sequence $0 \to I \to K^ a \to N \to 0$ we see that $I$ is a flat $R$-module. Thus $L^\bullet $ is a bounded above complex of flat modules, hence K-flat by Lemma 15.59.7. It follows that $E^\bullet $ is K-flat by Lemma 15.59.6. $\square$

Proof. Omitted. Hint: This just follows from the long exact cohomology sequence associated to a distinguished triangle and the fact that $- \otimes _ R^{\mathbf{L}} M$ preserves distinguished triangles. The easiest one to prove is (2) and the others follow from it by translation. $\square$

Proof. Assume (2). Then the complex $E^\bullet $ with $E^{-i} = F_ i$ is quasi-isomorphic to $M$. Hence the Tor dimension of $M$ is at most $d$ by Lemma 15.66.3. Conversely, assume (1). Let $P^\bullet \to M$ be a projective resolution of $M$. By Lemma 15.66.2 we see that $\tau _{\geq -d}P^\bullet $ is a flat resolution of $M$ of length $d$, i.e., (2) holds. $\square$

Proof. Clear from the fact that the Tor functors are additive. $\square$

Proof. Follows by induction on the length of the finite complex: use Lemma 15.66.5 and the stupid truncations. $\square$

Proof. Follows by induction on the length of the finite complex: use Lemma 15.66.5 and the canonical truncations. $\square$

Proof. We may assume that $K^\bullet $ is a complex of flat $B$-modules with $K^ i = 0$ for $i \not\in [a, b]$, see Lemma 15.66.3. Let $M$ be an $A$-module. Choose a free resolution $F^\bullet \to M$. Then

see Homology, Remark 12.18.4 for the second equality. By assumption (2) the complex $\text{Tot}(L^\bullet \otimes _ A F^\bullet )$ has nonzero cohomology only in degrees $[c, d]$. Hence the spectral sequence of Homology, Lemma 12.25.1 for the double complex $K^\bullet \otimes _ B \text{Tot}(L^\bullet \otimes _ A F^\bullet )$ proves that $(K^\bullet \otimes _ B^\mathbf {L} L^\bullet ) \otimes _ A^{\mathbf{L}} M$ has nonzero cohomology only in degrees $[a + c, b + d]$. $\square$

Proof. This is a special case of Lemma 15.66.10, but can also be seen directly as follows. We have $K^\bullet \otimes _ A^{\mathbf{L}} M = K^\bullet \otimes _ B^{\mathbf{L}} (M \otimes _ A B)$ since any projective resolution of $K^\bullet $ as a complex of $B$-modules is a flat resolution of $K^\bullet $ as a complex of $A$-modules and can be used to compute $K^\bullet \otimes _ A^{\mathbf{L}} M$. $\square$

Proof. This is a special case of Lemma 15.66.10, but can also be seen directly as follows. Let $M$ be an $A$-module. Choose a free resolution $F^\bullet \to M$. Then

By our assumption on $B$ as an $A$-module we see that $M \otimes _ A^{\mathbf{L}} B$ has cohomology only in degrees $-d, -d + 1, \ldots , 0$. Because $K^\bullet $ has tor amplitude in $[a, b]$ we see from the spectral sequence in Example 15.62.4 that $K^\bullet \otimes _ B^{\mathbf{L}} (M \otimes _ A^{\mathbf{L}} B)$ has cohomology only in degrees $[-d + a, b]$ as desired. $\square$

Proof. By Lemma 15.66.3 we can find a quasi-isomorphism $E^\bullet \to K^\bullet $ where $E^\bullet $ is a complex of flat $A$-modules with $E^ i = 0$ for $i \not\in [a, b]$. Then $E^\bullet \otimes _ A B$ computes $K^\bullet \otimes _ A ^{\mathbf{L}} B$ by construction and each $E^ i \otimes _ A B$ is a flat $B$-module by Algebra, Lemma 10.39.7. Hence we conclude by Lemma 15.66.3. $\square$

Proof. Immediate consequence of Lemma 15.66.13 and the fact that $M \otimes _ A^{\mathbf{L}} B = M \otimes _ A B$ because $B$ is flat over $A$. $\square$

Proof. Assume (3) and let $M$ be an $A$-module. Then $H^ i = H^ i(K^\bullet \otimes _ A^\mathbf {L} M)$ is a $B$-module and $(H^ i)_\mathfrak m = H^ i(K^\bullet _\mathfrak m \otimes _{A_\mathfrak p}^\mathbf {L} M_\mathfrak p)$. Hence $H^ i = 0$ for $i \not\in [a, b]$ by Algebra, Lemma 10.23.1. Thus (3) $\Rightarrow $ (1). We omit the proofs of (1) $\Rightarrow $ (2) and (2) $\Rightarrow $ (3). $\square$

Proof. This follows immediately from Lemma 15.66.15 but can also be seen directly as follows. Note that $- \otimes _ R R_{f_ i}$ is an exact functor and that therefore

and similarly for every $R$-module $M$ we have

Hence the result follows from the fact that an $R$-module $N$ is zero if and only if $N_{f_ i}$ is zero for each $i$, see Algebra, Lemma 10.23.2. $\square$

Proof. Let $M$ be an $R$-module. Since $R \to R'$ is flat we see that

and taking cohomology commutes with tensoring with $R'$. Hence $\text{Tor}_ i^ R(M, K^\bullet ) \otimes _ R R' = \text{Tor}_ i^{R'}(M \otimes _ R R', K^\bullet \otimes _ R R')$. Since $R \to R'$ is faithfully flat, the vanishing of $\text{Tor}_ i^{R'}(M \otimes _ R R', K^\bullet \otimes _ R R')$ for $i \not\in [a, b]$ implies the same thing for $\text{Tor}_ i^ R(M, K^\bullet )$. $\square$

Proof. This is true because for an $R$-module $M$ we have $H^ i(K \otimes _ R^\mathbf {L} M) \otimes _ A B = H^ i((K \otimes _ A^\mathbf {L} B) \otimes _ R^\mathbf {L} M)$ for all $i$. Namely, represent $K$ by a complex $K^\bullet $ of $A$-modules and choose a free resolution $F^\bullet \to M$. Then we have the equality

The cohomology groups of the left hand side are $H^ i((K \otimes _ A^\mathbf {L} B) \otimes _ R^\mathbf {L} M)$ and on the right hand side we obtain $H^ i(K \otimes _ R^\mathbf {L} M) \otimes _ A B$. $\square$

Proof. The assumption on $R$ means that every module has a finite projective resolution of length at most $d$, in particular every module has tor dimension $\leq d$. The second statement follows from Lemma 15.66.9 and the definitions. The third statement is a rephrasing of the second. $\square$

Proof. One direction follows from Lemma 15.66.13. For the other, assume $K$ has tor amplitude in $[a, b]$ and let $M'$ be an $R'$-module. We have to show that $K' \otimes _{R'}^\mathbf {L} M'$ has nonzero cohomology only for degrees contained in the interval $[a, b]$.

Let $I = \mathop{\mathrm{Ker}}(R' \to R)$. Then $I^ n = 0$ for some $n$. If $IM' = 0$, then we can view $M'$ as an $R$-module and argue as follows

which has nonvanishing cohomology only in the interval $[a, b]$ by assumption on $K$. If $I^{t + 1}M' = 0$, then we consider the short exact sequence

By induction on $t$ we have that both $K' \otimes _{R'}^\mathbf {L} IM'$ and $K' \otimes _{R'}^\mathbf {L} M'/IM'$ have nonzero cohomology only for degrees in the interval $[a, b]$. Then the distinguished triangle

proves the same is true for $K' \otimes _{R'}^\mathbf {L} M'$ as desired. $\square$