nLab Tor

Idea

In the context of homological algebra, the $Tor$-functor is the derived tensor product: the left derived functor of the tensor product of $R$-modules, for $R$ a commutative ring.

Together with the Ext-functor it constitutes one of the central operations of interest in homological algebra.

Definition

Given a ring $R$ the bifunctor $\otimes_R : Mod_{R} \times {}_{R}Mod\to Ab$ from two copies of $R$-Mod to Ab is a right exact functor. Its left derived functors are the Tor-functors

$$Tor(-,B) : Mod_R \to Ab$$

and

$$Tor(A,-) : {}_{R}Mod \to Ab$$

with respect to one argument with fixed another, if they exist, are parts of a bifunctor

$$Tor : Mod_{R}\times {}_{R}Mod\to Ab \,.$$

Properties

Existence and balancing

Given a right $R$-module

$$A \in Mod_R$$

and a left $R$-module

$$B \in {}_R Mod$$

there are in principle three different ways to compute their derived tensor product $Tor_\bullet(A,B)$:

1. keeping $B$ fixed and deriving the functor

   $$(-) \otimes_R B : Mod_R \to Ab$$

2. keeping $A$ fixed and deriving the functor

   $$A \otimes_R (-) : {}_R Mod \to Ab$$

3. deriving the functor

   $$(-) \otimes_R (-) : Mod_R \times {}_R Mod \to Ab$$

   in both arguments

Theorem

If both $Mod_{R}$ and $_{R}Mod$ have enough projectives, then all these three derived functors exist and all give the same result.

Proof

Existence is clear from the very definition of derived functor in homological algebra. So we show that deriving in the left argument gives the same result as deriving in the right argument.

Let $Q^A_\bullet \stackrel{\simeq_{qi}}{\to} A$ and $Q^B_\bullet \stackrel{\simeq_{qi}}{\to} B$ be projective resolutions of $A$ and $B$, respectively. The corresponding tensor product of chain complexes $Tot (Q^A_\bullet\otimes Q^B_\bullet)$, hence the total complex of the degreewise tensor product of modules double complex carries the filtration by horizontal degree as well as that by vertical degree.

Accordingly there are the corresponding two spectral sequences of a double complex, to be denoted here $\{{}^{A}E^r_{p,q}\}_{r,p,q}$ (for the filtering by $A$-degree) and $\{{}^{B}E^r_{p,q}\}_{r,p,q}$ (for the filtering by $B$-degree). By the discussion there, both converge to the chain homology of the total complex.

We find the value of both spectral sequences on low degree pages according to the general discussion at spectral sequence of a double complex - low degree pages.

The 0th page for both is

$${}^A E^0_{p,q} = {}^B E^0_{p,q} \coloneqq Q^A_p \otimes_R Q^B_q \,.$$

For the first page we have

$$\begin{aligned} {}^A E^1_{p,q} & \simeq H_q(C_{p,\bullet}) \\ & \simeq H_q( Q^A_p \otimes Q^B_\bullet ) \end{aligned}$$

and

$$\begin{aligned} {}^B E^1_{p,q} & \simeq H_q(C_{\bullet,p}) \\ & \simeq H_q( Q^A_\bullet \otimes Q^B_p ) \end{aligned} \,.$$

Now using the universal coefficient theorem in homology and the fact that $Q^A_\bullet$ and $Q^B_\bullet$ is a resolution by projective objects, by construction, hence of tensor acyclic objects for which all Tor-modules vanish, this simplifies to

$$\begin{aligned} {}^A E^1_{p,q} & \simeq Q^A_p \otimes H_q(Q^B_\bullet) \\ & \simeq \left\{ \array{ Q^A_p \otimes_R B & if\; q = 0 \\ 0 & otherwise } \right. \end{aligned}$$

and similarly

$$\begin{aligned} {}^B E^1_{p,q} & \simeq H_q(Q^A_\bullet) \otimes_R Q^B_p \\ & \simeq \left\{ \array{ A \otimes_R Q^B_p & if\; q = 0 \\ 0 & otherwise } \right. \end{aligned} \,.$$

It follows for the second pages that

$$\begin{aligned} {}^A E^2_{p,q} & \simeq H_p(H^{vert}_q(Q^A_\bullet \otimes Q^B_\bullet)) \\ & \simeq \left\{ \array{ (L_p( (-)\otimes_R B ))(A) & if \; q = 0 \\ 0 & otherwise } \right. \end{aligned}$$

and

$$\begin{aligned} {}^B E^2_{p,q} & \simeq H_p(H^{hor}_q(Q^A_\bullet \otimes Q^B_\bullet)) \\ & \simeq \left\{ \array{ (L_p ( A \otimes_R (-) ))(B) & if \; q = 0 \\ 0 \; otherwise } \right. \end{aligned} \,.$$

Now both of these second pages are concentrated in a single row and hence have converged on that page already. Therefore, since they both converge to the same value:

$$L_p((-)\otimes_R B)(A) \simeq {}^A E^2_{p,0} \simeq {}^A E^\infty_{p,0} \simeq {}^B E^2_{p,0} \simeq L_p(A \otimes_R (-))(B) \,.$$

Respect for direct sums and filtered colimits

Proposition

Each $Tor_n^R(-,N)$ respects direct sums.

Proof

Let $S \in$ Set and let $\{N_s\}_{s \in S}$ be an $S$-family of $R$-modules. Observe that

1. if $\{(F_s)_\bullet\}_{s \in S}$ is an family of projective resolutions, then their degreewise direct sum $(\oplus_{s \in S} F)_\bullet$ is a projective resolution of $\oplus_{s \in S} N_s$.

2. the tensor product functor distributes over direct sums (this is discussed at tensor product of modules – monoidal category structure)

3. the chain homology functor preserves direct sums (this is discussed at chain homology - respect for direct sums).

Using this we have

$$\begin{aligned} Tor_n^R(\oplus_{s \in S} N_s, N) & \simeq H_n\left( \left(\oplus_{s \in S} F\right) \otimes N \right) \\ & \simeq H_n\left( \oplus_{s \in S} \left(F_s \otimes N \right) \right) \\ & \simeq \oplus_{s \in S} H_n( F_s \otimes N ) \\ & \simeq \oplus_{s \in S} Tor_n(N_s, N) \end{aligned} \,.$$

Proposition

Each $Tor_n^R(-,N)$ respects filtered colimits.

Proof

Let hence $A \colon I \to R Mod$ be a filtered diagram of modules. For each $A_i$, $i \in I$ we may find a projective resolution and in fact a free resolution $(Y_i)_\bullet \stackrel{\simeq_{qi}}{\to} A$. Since chain homology commutes with filtered colimits (this is discussed at chain homology - respect for filtered colimits), this means that

$$(\underset{\to_i}{\lim} Y_i)_\bullet \to A$$

is still a quasi-isomorphism. Moreover, by Lazard's criterion the degreewise filtered colimits of free modules $\underset{\to_i}{\lim} (Y_i)_n$ for each $n \in \mathbb{N}$ are flat modules. This means that $\underset{\to_i}{\lim} (Y_i)_\bullet \to A$ is flat resolution of $A$. By the very definition or else by the basic properties of flat modules, this means that it is a $(-)\otimes N$-acyclic resolution. By the discussion there it follows that

$$Tor_n^\mathbb{Z}(A,N) \simeq H_n( (\underset{\to_i}{\lim} Y_i) \otimes N ) \,.$$

Now the tensor product of modules is a left adjoint functor (the right adjoint being the internal hom of modules) and so it commutes over the filtered colimit to yield, using again that chain homology commutes with filtered colimits,

$$\begin{aligned} \cdots & \simeq H_n( \underset{\to_i}{\lim} (Y_i \otimes N) ) \\ & \simeq \underset{\to_i}{\lim} H_n( Y_i \otimes N ) \\ & \simeq \underset{\to_i}{\lim} Tor_n( A_i, N) \end{aligned} \,.$$

Symmetry in the two arguments

Proposition

For $N_1, N_2 \in R Mod$ and $n \in \mathbb{N}$ there is a natural isomorphism

$$Tor_n(A,B) \simeq Tor_n(B,A) \,.$$

We first give a proof for $R$ a principal ideal domain such as $\mathbb{Z}$.

Proof

Let $R$ be a principal ideal domain such as $\mathbb{Z}$ (in the latter case $R$Mod$\simeq$ Ab). Then by the discussion at projective resolution – length-1 resolutions there is always a short exact sequence

$$0 \to F_1 \to F_0 \to N \to 0$$

exhibiting a projective resolution of any module $N$. It follows that $Tor_{n \geq 2}(-,-) = 0$.

Let then $0 \to F_1 \to F_2 \to N_2 \to 0$ be such a short resolution for $N_2$. Then by the long exact sequence of a derived functor this induces an exact sequence of the form

$$0 \to Tor_1(N_1, F_1) \to Tor_1(N_1, F_0) \to Tor_1(N_1, N_2) \to N_1 \otimes F_1 \to N_1 \otimes F_0 \to N_1 \otimes N_2 \to 0 \,.$$

Since by construction $F_0$ and $F_1$ are already projective modules themselves this collapses to an exact sequence

$$0 \to Tor_1(N_1, N_2) \hookrightarrow N_1 \otimes F_1 \to N_1 \otimes F_0 \to N_1 \otimes N_2 \to 0 \,.$$

To the last three terms we apply the natural symmetric braiding isomorphism in $(R Mod, \otimes_R)$ to get

$$\array{ 0 &\to& Tor_1(N_1, N_2) &\hookrightarrow& N_1 \otimes F_1 &\to& N_1 \otimes F_0 &\to& N_1 \otimes N_2 &\to& 0 \\ && \downarrow && \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{\simeq}} && \\ 0 &\to& Tor_1(N_2, N_1) &\hookrightarrow& F_1 \otimes N_1 &\to& F_0 \otimes N_1 &\to& N_2 \otimes N_1 &\to& 0 } \,.$$

This exhibits a morphism $Tor_1(N_1,N_2) \to Tor_1(N_2, N_1)$ as the morphism induced on kernels from an isomorphism between two morphisms. Hence this is itself an isomorphism. (This is just by the universal property of the kernel, but one may also think of it as a simple application of the the four lemma/five lemma.)

Localization

(…)

For instance (Weibel, cor. 3.2.13).

Explicit computations for $Tor_1$

Over a commutative ring $R$, the computation of the Tor functor can be reduced to the computation of each $Tor(M, R/I)$ where $I$ is a finitely generated ideal of $R$.

$Tor_1$ and the torsion modules

We shall start looking at the case where $I$ is a principal ideal generated by a regular element $r \in R$.

Definition

For $M$ and $R$-module and $r$ a regular element of $R$, we write

$${}_r M \coloneqq \{ x \in M | r \cdot x = 0 \}$$

for the $r$-torsion submodule?.

The following proposition explains why Tor functors are called this way.

Proposition

Let $R$ be a commutative ring, let $M$ be an $R$-module and let $r \in R$ be a regular element of $R$, then

$$Tor_1^R(R/(r), M) = {}_r M$$

Proof

Since $r$ is regular, one has a short exact sequence

$$0 \to R \stackrel{\cdot r}{\to} R \stackrel{mod\, r}{\to} R/(r) \to 0$$

which once tensored with $M$ gives us the long exact sequence

$$0 \to Tor_1(R/(r), M) \to M \stackrel{\cdot r}{\to} M \to M/rM \to 0$$

One can then identify $Tor_1(R/(r), M)$ with the kernel of the map $x \mapsto rx$, that is ${}_r M$.

This is no longer the case when $r$ is not regular.

Proposition

Let $\pi$ be an idempotent element of $R$, then the $R$-module $R/(\pi)$ is flat, that is

$$Tor_1^R(R/(\pi), X) = 0$$

for every $R$-module $X$.

Proof

Since $\pi$ is idempotent, $R/(\pi)$ can be identified with the direct summand $(1-\pi)R$ of $R$. It is then a projective $R$-module and is then flat.

The case of abelian groups

For $n \in \mathbb{N}$ with $n \geq 1$, write $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$ for the cyclic group of order $n$, as usual.

In the ring $\mathbb{Z}$ all non-zero elements are regular, we have thus that for any abelian group $A$

$$Tor_1^\mathbb{Z}(\mathbb{Z}_n, A) \simeq {}_n A$$

for every $n \neq 0$.

One can now leverage the knowledge of the structure of finite abelian groups to deduce the following proposition.

Proposition

Let $A$ be a finite abelian group and $B$ any abelian group. Then $Tor_1(A,B)$ is a torsion group. Specifically, $Tor_1(A,B)$ is a direct sum of torsion subgroups of $B$.

Proof

By a fundamental fact about finite abelian groups (see this theorem), $A$ is a direct sum of cyclic group $A \simeq \oplus_k \mathbb{Z}_{p_k}$. By prop. $Tor_1$ respects this direct sum, so that

$$Tor_1(A,B) \simeq \oplus_k Tor_1(\mathbb{Z}_{p_k}, B) \,.$$

By prop. every direct summand on the right is a torsion group and hence so is the whole direct sum.

More generally we have:

Proposition

Let $A$ and $B$ be abelian groups. Write $Tor^\mathbb{Z}$ for the left derived functor of tensoring over $R = \mathbb{Z}$. Then

1. $Tor^\mathbb{Z}_1(A,B)$ is a torsion group. Specifically it is a filtered colimit of torsion subgroups of $B$.

2. $Tor^{\mathbb{Z}}_1(\mathbb{Q}/\mathbb{Z}, A)$ is the torsion subgroup of $A$.

3. $A$ is a torsion-free group precisely if $Tor^\mathbb{Z}_1(A,-) = 0$, equivalently if $Tor^\mathbb{Z}_1(-,A) = 0$.

For instance (Weibel, prop. 3.1.2, prop. 3.1.3, cor. 3.1.5).

Proof

The group $A$ may be expressed as a filtered colimit

$$A \simeq \underset{\to_i}{\lim} A_i$$

of finitely generated subgroups (this is discussed at Mod - Limits and colimits). Each of these is a direct sum of cyclic groups.

By prop. $Tor_1^\mathbb{Z}(-,B)$ preserves these colimits. By prop. every cyclic group is sent to a torsion group (of either $A$ or $B$). Therefore by prop. $Tor_1(A,B)$ is a filtered colimit of direct sums of torsion groups. This is itself a torsion group.

Remark

Analogous results fail, in general, for $\mathbb{Z}$ replaced by another ring $R$.

Corollary

An abelian group is torsion free precisely if regarded as a $\mathbb{Z}$-module it is a flat module.

See at flat module - Examples for more.

$Tor_1$ of two cyclic modules

Proposition

Let $R$ be a commutative ring and let $I \subset R$ and $J \subset R$ be two ideals. Then

$$Tor^R_1(R/I, R/J) \simeq (I \cap J) /IJ$$

Proof

One has a short exact sequence $0 \to I \to R \to R/I \to 0$. Tensoring it with $R/J$, one gets the long exact sequence

$$0 \to Tor_1(R/I, R/J) \to I \otimes R/J \to R/J \to R/I \otimes R/J \to 0$$

Hence $Tor_1(R/I, R/J)$ can be identified with the kernel of the map $I \otimes R/J \to R/J$ which is isomorphic to $(I \cap J)/IJ$.

Related concepts

• Ext

• Cotor

• flat resolution lemma

• universal coefficient theorem

References

Standard textbook accounts include the following:

• Charles Weibel, An Introduction to Homological Algebra, Cambridge Studies in Adv. Math. 38, CUP 1994

• Henri Cartan, Samuel Eilenberg, Homological algebra, Princeton Univ. Press 1956.

• M. Kashiwara and P. Schapira, Categories and Sheaves, Springer (2000)

• S. I . Gelfand, Yu. I. Manin, Methods of homological algebra

Lecture notes include

• Daniel Murfet, Tor (pdf)

section 3 of

• Peter May, Notes on Tor and Ext (pdf)

and specifically for symmetric model categories of spectra

• Anthony Elmendorf, Igor Kriz, Peter May, section 7 of Modern foundations for stable homotopy theory (pdf)

Original articles include

• Patrick Keef, On the Tor functor and some classes of abelian groups, Pacific J. Math. Volume 132, Number 1 (1988), 63-84. (Euclid)