Exercise 3.2.3

Let R = k[x, y] and I be the maximal ideal (x, y) in R. we define two maps

\begin{matrix}R & ⟶^{d_{2}} & R ⊕ R \\ f & ⟼ & (−yf, xf) \end{matrix}

\begin{matrix}R ⊕ R & ⟶^{d_{1}} & R \\ (p, q) & ⟼ & xp + yq \end{matrix}

which are R-modules homomorphism. We shall show that we then have an exact sequence of R-modules (which is a projective resolution of k ≅ R/I as a R-module) :

0 ⟶ R ⟶^{d_{2}} R ⊕ R ⟶^{d_{1}} R ⟶^{d_{0}} k ⟶ 0 (1)

where d_{0} is the canonical projection. Clearly, d_{1}d_{2} = 0 (i.e. Imd_{2} ⊆ ker d_{1}). Let (p, q) ∈ ker d_{1}, then xp + yq = 0 in R implies that p = −yh and q = xh for some h ∈ R, i.e. (p, q) = d_{2}(h). Hence, Imd_{2} = ker d_{1}. We have Imd_{1} = I = ker d_{0}, d_{0} is onto and d_{2} is one-to-one. Therefore, (1) is an exact sequence. This projective resolution gives us the complex

C_{•} 0 ⟶ R ⟶^{d_{2}} R ⊕ R ⟶^{d_{1}} R ⟶ 0 .

Applying the functor k ⊗_{R} − to C_{•} gives the complex

k ⊗_{R} C_{•} 0 ⟶ k ⊗_{R} R ⟶^{k⊗d_{2}} k ⊗_{R} (R ⊕ R) ⟶^{k⊗d_{1}} k ⊗_{R} R ⟶ 0 .

Then Tor_{2}^{R}(k, k) = H_{2}(k ⊗_{R} C_{•}) = ker(k ⊗ d_{2}). We have k ⊗_{R} R ≅ k and k ⊗_{R} (R ⊕ R) ≅ k ⊕ k. With respect to these identifications λ ⊗ f in k ⊗_{R} R becomes λf(0, 0) in k. Therefore, (k⊗d_{2})(λ⊗f) = (−λ⊗yf, λ⊗xf) = (0, 0) in k ⊕ k. This means that the map k ⊗ d_{2} is the nul map, since it is 0 on the generators, hence ker(k ⊗ d_{2}) = k. We shall now show that we have a R-module isomorphism Tor_{2}^{R}(k, k) ≅ Tor_{1}^{R}(I, k). We have a short exact sequence

0 ⟶ I ⟶ R ⟶ k ⟶ 0

which gives a long exact sequence of homology

⋯ ⟶ \underbrace{Tor_{2}^{R}(R, k)}_{=0} ⟶ Tor_{2}^{R}(k, k) ⟶^{δ} Tor_{1}^{R}(I, k) ⟶ \underbrace{Tor_{1}^{R}(R, k)}_{=0} ⟶ ⋯

i.e. δ is an isomorphism.