Exercise B17 (Weibel, Exercise 3.1.3, pg. 68)

Show that Tor_{1}^{R}(R/I, R/J) ≅ \frac{I∩J}{IJ} for every right ideal I and every left ideal J of R. In particular Tor_{1}^{R}(R/I, R/I) ≅ I/I^{2}.

Proof. To prove this result, we need the following lemma.

Lemma 1. For every right ideal I and every left ideal J of R, the following diagram is a commutative diagram:

0 → IJ → I → I ⊗ R/J → 0
↓^{f_{1}} ↓^{f_{2}} ↓^{f_{3}}
0 → J → R → R ⊗ R/J → 0

where f_{1} and f_{2} are inclusion maps, and f_{3} = f_{2} ⊗ id_{R/J}

Proof. We see that the top and bottom rows of the above diagram are short exact sequences once we observe that I ⊗_{R} R/J ≅ I/IJ and R ⊗ R/J ≅ R/J. It is also easy to see that the first square commutes. The second square commutes because the maps I → I ⊗ R/J and R → R ⊗ R/J are defined to be f ↦ f ⊗ 1.

From the short exact sequence of R-modules

0 → I → R → R/I → 0

we have the following long exact sequence

⋯ → Tor_{1}^{R}(R, R/J) → Tor_{1}^{R}(R/I, R/J) → Tor_{0}^{R}(I, R/J) →^{d}

Tor_{0}^{R}(R, R/J) → Tor_{0}^{R}(R/I, R/J) → 0

where d(f ⊗\overline{g}) = f ⊗\overline{g}, where f ∈ I and \overline{g} ∈ R/J. Since R is an R-projective module, Tor_{1}^{R}(R, R/J) = 0. Hence,

Tor_{1}^{R}(R/I, R/J) ≅ ker(Tor_{0}^{R}(I, R/J) → Tor_{0}^{R}(R, R/J))

Since Tor_{0}^{R}(A, B) = A ⊗ B for any two R-modules A and B, we need to calculate

Tor_{1}^{R}(R/I, R/J) ≅ ker(I ⊗ R/J →^{d} R ⊗ R/J) = ker(f_{3})

where f_{3} is the map in our lemma. By applying the snake lemma to the commutative diagram in the above lemma we have the following exact sequence:

0 → ker(f_{3}) → coker(f_{1}) = J/IJ → coker(f_{2}) = R/I → coker(f_{3}) → 0.

Hence

ker(f_{3}) ≅ ker(J/IJ → R/I).

A routine check will now show that

ker(f_{3}) ≅ \frac{I ∩ J}{IJ}

which is what we wanted to show.