(also nonabelian homological algebra)
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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.
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
and
with respect to one argument with fixed another, if they exist, are parts of a bifunctor
Given a right $R$-module
and a left $R$-module
there are in principle three different ways to compute their derived tensor product $Tor_\bullet(A,B)$:
keeping $B$ fixed and deriving the functor
keeping $A$ fixed and deriving the functor
deriving the functor
in both arguments
If both $Mod_{R}$ and $_{R}Mod$ have enough projectives, then all these three derived functors exist and all give the same result.
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
For the first page we have
and
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
and similarly
It follows for the second pages that
and
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:
Each $Tor_n^R(-,N)$ respects direct sums.
Let $S \in$ Set and let $\{N_s\}_{s \in S}$ be an $S$-family of $R$-modules. Observe that
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 $\oplus_{s \in S} N_s$.
the tensor product functor distributes over direct sums (this is discussed at tensor product of modules – monoidal category structure)
the chain homology functor preserves direct sums (this is discussed at chain homology - respect for direct sums).
Using this we have
Each $Tor_n^R(-,N)$ respects filtered colimits.
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
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
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,
For $N_1, N_2 \in R Mod$ and $n \in \mathbb{N}$ there is a natural isomorphism
We first give a proof for $R$ a principal ideal domain such as $\mathbb{Z}$.
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
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
Since by construction $F_0$ and $F_1$ are already projective modules themselves this collapses to an exact sequence
To the last three terms we apply the natural symmetric braiding isomorphism in $(R Mod, \otimes_R)$ to get
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.)
(…)
For instance (Weibel, cor. 3.2.13).
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 $R$.
We shall start looking at the case where $I$ is a principal ideal generated by a regular element $r \in R$.
For $M$ and $R$-module and $r$ a regular element of $R$, we write
for the $r$-torsion submodule?.
The following proposition explains why Tor functors are called this way.
Let $R$ be a commutative ring, let $M$ be an $R$-module and let $r \in R$ be a regular element of $R$, then
Since $r$ is regular, one has a short exact sequence
which once tensored with $M$ gives us the long exact sequence
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.
Let $\pi$ be an idempotent element of $R$, then the $R$-module $R/(\pi)$ is flat, that is
for every $R$-module $X$.
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.
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 $A$
for every $n \neq 0$.
One can now leverage the knowledge of the structure of finite abelian groups to deduce the following 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 $B$.
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
By prop. every direct summand on the right is a torsion group and hence so is the whole direct sum.
More generally we have:
Let $A$ and $B$ be abelian groups. Write $Tor^\mathbb{Z}$ for the left derived functor of tensoring over $R = \mathbb{Z}$. Then
$Tor^\mathbb{Z}_1(A,B)$ is a torsion group. Specifically it is a filtered colimit of torsion subgroups of $B$.
$Tor^{\mathbb{Z}}_1(\mathbb{Q}/\mathbb{Z}, A)$ is the torsion subgroup of $A$.
$A$ 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).
The group $A$ may be expressed as a filtered colimit
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.
Analogous results fail, in general, for $\mathbb{Z}$ replaced by another ring $R$.
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.
Let $R$ be a commutative ring and let $I \subset R$ and $J \subset R$ be two ideals. Then
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
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 $(I \cap J)/IJ$.
Standard textbook accounts include the following:
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
section 3 of
and specifically for symmetric model categories of spectra
Original articles include
Last revised on July 18, 2023 at 22:20:05. See the history of this page for a list of all contributions to it.