Exercise 3.1.1

Let R be a commutative ring and B a R-module. Let r ∈ R, we denote by _{r}R = {x ∈ R : rx = 0} and _{r}B = {b ∈ B : rb = 0}. _{r}R is an ideal of R and _{r}B is a submodule of B. We denote by rR the ideal of R generated by r. We have a well defined R-linear map

\begin{matrix}_{r}R ⊗_{R} B & ⟶^{μ} & _{r}B \\ x ⊗ b & ⟼ & xb \end{matrix}

with Imμ = _{r}RB = {∑_{k} x_{k}b_{k} : (x_{k}, b_{k}) ∈ _{r}R × B}, which gives an exact sequence of R-modules

0 ⟶ ker μ ⟶ _{r}R ⊗_{R} B ⟶^{μ} _{r}B ⟶ _{r}B/_{r}RB ⟶ 0 .

We shall show that Tor_{1}^{R}(R/rR, B) ≅ _{r}B/_{r}RB and Tor_{2}^{R}(R/rR, B) ≅ ker μ. With a short exact sequence 0 ⟶ rR ⟶ R ⟶ R/rR ⟶ 0 we have a long exact sequence of homology

⋯ ⟶ \underbrace{Tor_{2}^{R}(R, B)}_{=0} ⟶ Tor_{2}^{R}(R/rR, B) ⟶^{δ} Tor_{1}^{R}(rR, B) ⟶ \underbrace{Tor_{1}^{R}(R, B)}_{=0} ⟶

⟶ Tor_{1}^{R}(R/rR, B) ⟶^{δ} rR ⊗_{R} B ⟶^{φ} B ⟶ R/rR ⊗_{R} B ⟶ 0

which gives the R-modules isomorphisms Tor_{2}^{R}(R/rR, B) ≅ Tor_{1}^{R}(rR, B) and Tor_{1}^{R}(R/rR, B) ≅ ker φ. Let x = ∑_{k} rx_{k} ⊗ b_{k} ∈ rR ⊗_{R} B. Then φ(x) = r ∑_{k} x_{k}b_{k} = 0 iff ∑_{k} x_{k}b_{k} ∈ _{r}B. Since x = ∑_{k} rx_{k} ⊗ b_{k} = r ⊗ ∑_{k} x_{k}b_{k}, we have ker φ = {r ⊗ b : b ∈ _{r}B}. Then we have a projection _{r}B ⟶^{π} ker φ s.t. π(b) = r ⊗ b with r ⊗ b = 0 iff b = ∑_{k} x_{k}b_{k} for some x_{k} ∈ _{r}R and b_{k} ∈ B. Hence ker φ ≅ _{r}B/_{r}RB. Now with the short exact sequence 0 ⟶ _{r}R ⟶ R ⟶ rR ⟶ 0 we get a long exact sequence of homology

⋯ ⟶ \underbrace{Tor_{1}^{R}(R, B)}_{=0} ⟶ Tor_{1}^{R}(rR, B) ⟶^{δ} _{r}R⊗_{R}B ⟶^{ψ} B ⟶^{π} \underbrace{rR ⊗_{R} B}_{≅rB} ⟶ 0

so that Tor_{1}^{R}(rR, B) = ker ψ. Looking at the commutative diagram

0 ⟶ Tor_{1}^{R}(rR, B) ⟶ _{r}R ⊗_{R} B ⟶^{ψ} B ⟶^{π} rB ⟶ 0
μ↓  ↓^{id}  ↓^{id}
0  ⟶ _{r}B ⟶^{i} B ⟶^{π} rB ⟶ 0

with exacts rows gives us Imψ = ker π = Imi = _{r}B, i being the natural injection of _{r}B into B. Therefore ker μ = ker ψ = Tor_{1}^{R}(rR, B). But since Tor_{1}^{R}(rR, B) ≅ Tor_{2}^{R}(R/rR, B), we have the desired result. To show that Tor_{n}^{R}(R/rR, B) ≅ Tor_{n−2}^{R}(_{r}R, B) ∀n ≥ 3, first observe that the long exact sequence of homology associated with the short exact sequence 0 ⟶ rR ⟶ R ⟶ R/rR ⟶ 0 gives us the isomorphisms Tor_{n}^{R}(R/rR, B) ≅ Tor_{n−1}^{R}(rR, B) ∀n ≥ 2. Then the long exact sequence of homology associated with the short exact sequence 0 ⟶ _{r}R ⟶ R ⟶ rR ⟶ 0 gives us the isomorphisms Tor_{n−1}^{R}(rR, B) ≅ Tor_{n−2}^{R}(_{r}R, B) ∀n ≥ 3. Each isomorphism is due to the fact that Tor_{n}^{R}(R, −) = 0 ∀n ≥ 1.