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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.
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) = A ⊗R 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
i.e., we take a projective resolution
then remove the A term and tensor the projective resolution with B to get the complex
(note that A ⊗R B does not appear and the last arrow is just the zero map) and take the homology of this complex.
Properties
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.
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".
n(F, B) = 0 for all n ≥ 1.
1(A, B) whenever A is finitely generated.
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).
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
i and K
i are free abelian groups. This gives rise to a double-complex with exact rows and columnsStart with x ∈ TorZ
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). Theni.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. Theni.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.