## 15.66 Tor dimension

Instead of resolving by projective modules we can look at resolutions by flat modules. This leads to the following concept.

Definition 15.66.1. Let $R$ be a ring. Denote $D(R)$ its derived category. Let $a, b \in \mathbf{Z}$.

An object $K^\bullet $ of $D(R)$ has *tor-amplitude in $[a, b]$* if $H^ i(K^\bullet \otimes _ R^\mathbf {L} M) = 0$ for all $R$-modules $M$ and all $i \not\in [a, b]$.

An object $K^\bullet $ of $D(R)$ has *finite tor dimension* if it has tor-amplitude in $[a, b]$ for some $a, b$.

An $R$-module $M$ has *tor dimension $\leq d$* if $M[0]$ as an object of $D(R)$ has tor-amplitude in $[-d, 0]$.

An $R$-module $M$ has *finite tor dimension* if $M[0]$ as an object of $D(R)$ has finite tor dimension.

We observe that if $K^\bullet $ has finite tor dimension, then $K^\bullet \in D^ b(R)$.

Lemma 15.66.2. Let $R$ be a ring. Let $K^\bullet $ be a bounded above complex of flat $R$-modules with tor-amplitude in $[a, b]$. Then $\mathop{\mathrm{Coker}}(d_ K^{a - 1})$ is a flat $R$-module.

**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

\[ K^{a - 2} \otimes _ R M \to K^{a - 1} \otimes _ R M \to K^ a \otimes _ R M \]

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.
$\square$

Lemma 15.66.3. Let $R$ be a ring. Let $K^\bullet $ be an object of $D(R)$. Let $a, b \in \mathbf{Z}$. The following are equivalent

$K^\bullet $ has tor-amplitude in $[a, b]$.

$K^\bullet $ is quasi-isomorphic to a complex $E^\bullet $ of flat $R$-modules with $E^ i = 0$ for $i \not\in [a, b]$.

**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.
$\square$

Lemma 15.66.4. Let $R$ be a ring. Let $a \in \mathbf{Z}$ and let $K$ be an object of $D(R)$. The following are equivalent

$K$ has tor-amplitude in $[a, \infty ]$, and

$K$ is quasi-isomorphic to a K-flat complex $E^\bullet $ whose terms are flat $R$-modules with $E^ i = 0$ for $i \not\in [a, \infty ]$.

**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

\[ K^{n - 1} \otimes _ R M \to K^ n \otimes _ R M \to K^{n + 1} \otimes _ R M \]

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

\[ E^\bullet = \tau _{\geq a}K^\bullet = (\ldots \to 0 \to N \to K^{a + 1} \to \ldots ) \]

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

\[ L^\bullet = (\ldots \to K^{a - 1} \to I \to 0 \to \ldots ) \]

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.
$\square$

Lemma 15.66.5. Let $R$ be a ring. Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle in $D(R)$. Let $a, b \in \mathbf{Z}$.

If $K^\bullet $ has tor-amplitude in $[a + 1, b + 1]$ and $L^\bullet $ has tor-amplitude in $[a, b]$ then $M^\bullet $ has tor-amplitude in $[a, b]$.

If $K^\bullet , M^\bullet $ have tor-amplitude in $[a, b]$, then $L^\bullet $ has tor-amplitude in $[a, b]$.

If $L^\bullet $ has tor-amplitude in $[a + 1, b + 1]$ and $M^\bullet $ has tor-amplitude in $[a, b]$, then $K^\bullet $ has tor-amplitude in $[a + 1, b + 1]$.

**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.
$\square$

Lemma 15.66.6. Let $R$ be a ring. Let $M$ be an $R$-module. Let $d \geq 0$. The following are equivalent

$M$ has tor dimension $\leq d$, and

there exists a resolution

\[ 0 \to F_ d \to \ldots \to F_1 \to F_0 \to M \to 0 \]

with $F_ i$ a flat $R$-module.

In particular an $R$-module has tor dimension $0$ if and only if it is a flat $R$-module.

**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.
$\square$

Lemma 15.66.7. Let $R$ be a ring. Let $a, b \in \mathbf{Z}$. If $K^\bullet \oplus L^\bullet $ has tor amplitude in $[a, b]$ so do $K^\bullet $ and $L^\bullet $.

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

Lemma 15.66.8. Let $R$ be a ring. Let $K^\bullet $ be a bounded complex of $R$-modules such that $K^ i$ has tor amplitude in $[a - i, b - i]$ for all $i$. Then $K^\bullet $ has tor amplitude in $[a, b]$. In particular if $K^\bullet $ is a finite complex of $R$-modules of finite tor dimension, then $K^\bullet $ has finite tor dimension.

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

Lemma 15.66.9. Let $R$ be a ring. Let $a, b \in \mathbf{Z}$. Let $K^\bullet \in D^ b(R)$ such that $H^ i(K^\bullet )$ has tor amplitude in $[a - i, b - i]$ for all $i$. Then $K^\bullet $ has tor amplitude in $[a, b]$. In particular if $K^\bullet \in D^ b(R)$ and all its cohomology groups have finite tor dimension then $K^\bullet $ has finite tor dimension.

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

Lemma 15.66.10. Let $A \to B$ be a ring map. Let $K^\bullet $ and $L^\bullet $ be complexes of $B$-modules. Let $a, b, c, d \in \mathbf{Z}$. If

$K^\bullet $ as a complex of $B$-modules has tor amplitude in $[a, b]$,

$L^\bullet $ as a complex of $A$-modules has tor amplitude in $[c, d]$,

then $K^\bullet \otimes ^\mathbf {L}_ B L^\bullet $ as a complex of $A$-modules has tor amplitude in $[a + c, b + d]$.

**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

\[ (K^\bullet \otimes _ B^\mathbf {L} L^\bullet ) \otimes _ A^{\mathbf{L}} M = \text{Tot}(\text{Tot}(K^\bullet \otimes _ B L^\bullet ) \otimes _ A F^\bullet ) = \text{Tot}(K^\bullet \otimes _ B \text{Tot}(L^\bullet \otimes _ A F^\bullet )) \]

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 $[a + c, b + d]$.
$\square$

Lemma 15.66.11. Let $A \to B$ be a ring map. Assume that $B$ is flat as an $A$-module. Let $K^\bullet $ be a complex of $B$-modules. Let $a, b \in \mathbf{Z}$. If $K^\bullet $ as a complex of $B$-modules has tor amplitude in $[a, b]$, then $K^\bullet $ as a complex of $A$-modules has tor amplitude in $[a, b]$.

**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 $K^\bullet \otimes _ A^{\mathbf{L}} M$.
$\square$

Lemma 15.66.12. Let $A \to B$ be a ring map. Assume that $B$ has tor dimension $\leq d$ as an $A$-module. Let $K^\bullet $ be a complex of $B$-modules. Let $a, b \in \mathbf{Z}$. If $K^\bullet $ as a complex of $B$-modules has tor amplitude in $[a, b]$, then $K^\bullet $ as a complex of $A$-modules has tor amplitude in $[a - d, b]$.

**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

\[ K^\bullet \otimes _ A^{\mathbf{L}} M = \text{Tot}(K^\bullet \otimes _ A F^\bullet ) = \text{Tot}(K^\bullet \otimes _ B (F^\bullet \otimes _ A B)) = K^\bullet \otimes _ B^{\mathbf{L}} (M \otimes _ A^{\mathbf{L}} B). \]

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.
$\square$

Lemma 15.66.13. Let $A \to B$ be a ring map. Let $a, b \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $A$-modules with tor amplitude in $[a, b]$. Then $K^\bullet \otimes _ A^{\mathbf{L}} B$ as a complex of $B$-modules has tor amplitude in $[a, b]$.

**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.
$\square$

Lemma 15.66.14. Let $A \to B$ be a flat ring map. Let $d \geq 0$. Let $M$ be an $A$-module of tor dimension $\leq d$. Then $M \otimes _ A B$ is a $B$-module of tor dimension $\leq d$.

**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 $A$.
$\square$

Lemma 15.66.15. Let $A \to B$ be a ring map. Let $K^\bullet $ be a complex of $B$-modules. Let $a, b \in \mathbf{Z}$. The following are equivalent

$K^\bullet $ has tor amplitude in $[a, b]$ as a complex of $A$-modules,

$K^\bullet _\mathfrak q$ has tor amplitude in $[a, b]$ as a complex of $A_\mathfrak p$-modules for every prime $\mathfrak q \subset B$ with $\mathfrak p = A \cap \mathfrak q$,

$K^\bullet _\mathfrak m$ has tor amplitude in $[a, b]$ as a complex of $A_\mathfrak p$-modules for every maximal ideal $\mathfrak m \subset B$ with $\mathfrak p = A \cap \mathfrak m$.

**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).
$\square$

Lemma 15.66.16. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ be elements which generate the unit ideal. Let $a, b \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $R$-modules. If for each $i$ the complex $K^\bullet \otimes _ R R_{f_ i}$ has tor amplitude in $[a, b]$, then $K^\bullet $ has tor amplitude in $[a, b]$.

**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

\[ H^ i(K^\bullet )_{f_ i} = H^ i(K^\bullet ) \otimes _ R R_{f_ i} = H^ i(K^\bullet \otimes _ R R_{f_ i}). \]

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

\[ H^ i(K^\bullet \otimes _ R^{\mathbf{L}} M)_{f_ i} = H^ i(K^\bullet \otimes _ R^{\mathbf{L}} M) \otimes _ R R_{f_ i} = H^ i(K^\bullet \otimes _ R R_{f_ i} \otimes _{R_{f_ i}}^{\mathbf{L}} M_{f_ i}). \]

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.
$\square$

Lemma 15.66.17. Let $R$ be a ring. Let $a, b \in \mathbf{Z}$. Let $K^\bullet $ be a complex of $R$-modules. Let $R \to R'$ be a faithfully flat ring map. If the complex $K^\bullet \otimes _ R R'$ has tor amplitude in $[a, b]$, then $K^\bullet $ has tor amplitude in $[a, b]$.

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

\[ (M \otimes _ R^{\mathbf{L}} K^\bullet ) \otimes _ R R' = ((M \otimes _ R R') \otimes _{R'}^{\mathbf{L}} (K^\bullet \otimes _ R R') \]

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 $\text{Tor}_ i^ R(M, K^\bullet )$.
$\square$

Lemma 15.66.18. Given ring maps $R \to A \to B$ with $A \to B$ faithfully flat and $K \in D(A)$ the tor amplitude of $K$ over $R$ is the same as the tor amplitude of $K \otimes _ A^\mathbf {L} B$ over $R$.

**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

\[ \text{Tot}(K^\bullet \otimes _ A B \otimes _ R F^\bullet ) = \text{Tot}(K^\bullet \otimes _ R F^\bullet ) \otimes _ A B \]

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 $H^ i(K \otimes _ R^\mathbf {L} M) \otimes _ A B$.
$\square$

Lemma 15.66.19. Let $R$ be a ring of finite global dimension $d$. Then

every module has tor dimension $\leq d$,

a complex of $R$-modules $K^\bullet $ with $H^ i(K^\bullet ) \not= 0$ only if $i \in [a, b]$ has tor amplitude in $[a - d, b]$, and

a complex of $R$-modules $K^\bullet $ has finite tor dimension if and only if $K^\bullet \in D^ b(R)$.

**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.
$\square$

Lemma 15.66.20. Let $R' \to R$ be a surjective ring map whose kernel is a nilpotent ideal. Let $K' \in D(R')$ and set $K = K' \otimes _{R'}^\mathbf {L} R$. Let $a, b \in \mathbf{Z}$. Then $K$ has tor amplitude in $[a, b]$ if and only if $K'$ does.

**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

\[ K' \otimes _{R'}^\mathbf {L} M' = K' \otimes _{R'}^\mathbf {L} (R \otimes _ R^\mathbf {L} M') = (K' \otimes _{R'}^\mathbf {L} R) \otimes _ R^\mathbf {L} M' = K \otimes _ R^\mathbf {L} M' \]

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

\[ 0 \to IM' \to M' \to M'/IM' \to 0 \]

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

\[ K' \otimes _{R'}^\mathbf {L} IM' \to K' \otimes _{R'}^\mathbf {L} M' K' \otimes _{R'}^\mathbf {L} M'/IM' (K' \otimes _{R'}^\mathbf {L} IM')[1] \]

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

## Comments (2)

Comment #2148 by Juanyong Wang on

Comment #2184 by Johan on