# Tor functor

In homological algebra, the Tor functors are the derived functors of the tensor product of modules over a ring. They were first defined in general to express the Künneth theorem and universal coefficient theorem in algebraic topology.[citation needed] For abelian groups, Tor is related to the notion of a torsion subgroup, which explains the notation.

## Definition

Specifically, suppose R is a ring, and denote by R-Mod the category of left R-modules and by Mod-R the category of right R-modules (if R is commutative, the two categories coincide). Pick a fixed module B in R-Mod. For A in Mod-R, set T(A) = AR B. Then T is a right exact functor from Mod-R to the category of abelian groups Ab (in the case when R is commutative, it is a right exact functor from Mod-R to Mod-R) and its left derived functors LnT are defined. We set

${\displaystyle \mathrm {Tor} _{n}^{R}(A,B)=(L_{n}T)(A)}$

i.e., we take a projective resolution

${\displaystyle \cdots \to P_{2}\to P_{1}\to P_{0}\to A\to 0}$

then remove the A term and tensor the projective resolution with B to get the complex

${\displaystyle \cdots \to P_{2}\otimes _{R}B\to P_{1}\otimes _{R}B\to P_{0}\otimes _{R}B\to 0}$

(note that AR B does not appear and the last arrow is just the zero map) and take the homology of this complex.

## Properties

• For every n ≥ 1, TorR
n
is an additive functor from Mod-R × R-Mod to Ab. In the case when R is commutative, we have additive functors from Mod-R × Mod-R to Mod-R.
• As is true for every family of derived functors, every short exact sequence 0 → KLM → 0 induces a long exact sequence of the form
{\displaystyle {\begin{aligned}\cdots \to \mathrm {Tor} _{2}^{R}(M,B)\to \mathrm {Tor} _{1}^{R}(K,B)&\to \mathrm {Tor} _{1}^{R}(L,B)\to \mathrm {Tor} _{1}^{R}(M,B)\\[4pt]&\to K\otimes B\to L\otimes B\to M\otimes B\to 0.\end{aligned}}}
${\displaystyle \mathrm {Tor} _{1}^{R}(R/(r),B)=\{b\in B:rb=0\},}$
from which the terminology Tor (that is, Torsion) comes: see torsion subgroup.
• TorZ
n
(A, B) = 0 for all n ≥ 2. The reason: every abelian group A has a free resolution of length 1, since subgroups of free abelian groups are free abelian. So in this important special case, the higher Tor functors vanish. In addition, TorZ
1
(Z/kZ, A) = ker(f) where f: A → A represents "multiplication by k".
• Furthermore, every free module has a free resolution of length zero, so by the argument above, if F is a free R-module, then TorR
n
(F, B) = 0 for all n ≥ 1.
• The Tor functors preserve filtered colimits and arbitrary direct sums: there is a natural isomorphism
${\displaystyle \mathrm {Tor} _{n}^{R}\left(\bigoplus _{i}A_{i},\bigoplus _{j}B_{j}\right)\simeq \bigoplus _{i}\bigoplus _{j}\mathrm {Tor} _{n}^{R}(A_{i},B_{j})}$
• From the classification of finitely generated abelian groups, we know that every finitely generated abelian group is the direct sum of copies of Z and Zk. This together with the previous three points allows us to compute TorZ
1
(A, B) whenever A is finitely generated.
• A module M in Mod-R is flat if and only if TorR
1
(M, –) = 0. In this case, we even have TorR
n
(M, –) = 0 for all n ≥ 1. In fact, to compute TorR
n
(A, B), one may use a flat resolution of A or B, instead of a projective resolution (note that a projective resolution is automatically a flat resolution, but the converse isn't true, so allowing flat resolutions is more flexible).
• Symmetry: if R is commutative, then there is a natural isomorphism TorR
n
(L
1
, L
2
) ≅ TorR
n
(L
2
, L
1
). Here is the idea for abelian groups (i.e., the case R = Z and n = 1). Fix a free resolution of L
i
as follows
${\displaystyle 0\to M_{i}\to K_{i}\to L_{i}\to 0,}$
so that M
i
and K
i
are free abelian groups. This gives rise to a double-complex with exact rows and columns
1
(L
1
, L
2
), so β03(x) ∈ ker(β13). Let x
12
M
1
K
2
be such that α12(x
12
) = β03(x). Then
${\displaystyle \alpha _{22}\circ \beta _{12}(x_{12})=\beta _{13}\circ \alpha _{21}(x_{12})=\beta _{13}\circ \beta _{03}(x)=0,}$
i.e., β12(x
12
) ∈ ker(α22). By exactness of the second row, this means that β12(x
12
) = α21(x
21
) for some unique x
21
K
1
M
2
. Then
${\displaystyle \alpha _{31}\circ \beta _{21}(x_{21})=\beta _{22}\circ \alpha _{21}(x_{21})=\beta _{22}\circ \beta _{12}(x_{12})=0,}$
i.e., β21(x
21
) ∈ ker(α31). By exactness of the bottom row, this means that β21(x
21
) = α30(y) for some unique y ∈ TorZ
1
(L
2
, L
1
).
Upon checking that y is uniquely determined by x (not dependent on the choice of x
12
), this defines a function TorZ
n
(L
1
, L
2
) → TorZ
n
(L
2
, L
1
), taking x to y, which is a group homomorphism. One may check that this map has an inverse, namely the function TorZ
n
(L
2
, L
1
) → TorZ
n
(L
1
, L
2
) defined in a similarly manner. One can also check that the function does not depend on the choice of free resolutions.