# Tor functor

In homological algebra, the Tor functors are the derived functors of the tensor product functor. They were first defined in generality to express the Künneth theorem and universal coefficient theorem in algebraic topology.[citation needed]

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) = ARB. 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

$\mathrm{Tor}_n^R(A,B)=(L_nT)(A)$

i.e., we take a projective resolution

$\cdots\rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow A\rightarrow 0$

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

$\cdots \rightarrow P_2\otimes_R B \rightarrow P_1\otimes_R B \rightarrow P_0\otimes_R B \rightarrow 0$

(note that ARB 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.
$\cdots\rightarrow\mathrm{Tor}_2^R(M,B)\rightarrow\mathrm{Tor}_1^R(K,B)\rightarrow\mathrm{Tor}_1^R(L,B)\rightarrow\mathrm{Tor}_1^R(M,B)\rightarrow K\otimes B\rightarrow L\otimes B\rightarrow M\otimes B\rightarrow 0$.
$\mathrm{Tor}_1^R(R/(r),B)=\{b\in B:rb=0\},$

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 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.
$\mathrm{Tor}_n^R \left (\bigoplus_i A_i, \bigoplus_j B_j \right) \simeq \bigoplus_i \bigoplus_j \mathrm{Tor}_n^R(A_i,B_j)$
• 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
) $\simeq$ 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

$0 \to M_i \to K_i \to L_i,$

so that M
i
and K
i
are free abelian groups. This gives rise to a double-complex with exact rows and columns

Start with x $\in$ TorZ
1
(L
1
, L
2
), so $\beta_{03}$(x) $\in$ Ker($\beta_{13}$). Let x
12
$\in$ M
1
$\otimes$ K
2
be such that $\alpha_{12}$(x
12
) = $\beta_{03}$(x). Then

$\alpha_{22}\circ \beta_{12}(x_{12}) = \beta_{13} \circ \alpha_{21}(x_{12}) = \beta_{13} \circ \beta_{03}(x) = 0,$

i.e., $\beta_{12}$(x
12
) $\in$ Ker($\alpha_{22}$). By exactness of the second row, this means that $\beta_{12}$(x
12
) = $\alpha_{21}$(x
21
) for some unique x
21
$\in$ K
1
$\otimes$ M
2
. Then

$\alpha_{31}\circ\beta_{21}(x_{21}) = \beta_{22}\circ\alpha_{21}(x_{21}) = \beta_{22}\circ\beta_{12}(x_{12})=0,$

i.e., $\beta_{21}$(x
21
) $\in$ Ker($\alpha_{31}$). By exactness of the bottom row, this means that $\beta_{21}$(x
21
) = $\alpha_{30}$(y) for some unique y $\in$ 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
) $\to$ 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
) $\to$ 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.