I want to prove that $\operatorname{Tor}_{n}^{R}(k,k)=k\oplus k,\,\,\forall n\ge 1$.
I found the projective resolution $$ R^4\stackrel{d_2} \longrightarrow R^3\stackrel{d_1} \longrightarrow R^2\stackrel{d_0} \longrightarrow R \longrightarrow k $$ where $d_{0}(f,g)=xf+yg,\,\,\,\,d_{1}(f,g,h)=f(x,0)+g(y,0)+h(y,-x),\,\,\\ d_{2}(f,g,h,k)= f(y,x,0)+g(x+y,0,x)+h(0,x-y,y)+k(1,-1,1).$
But i can't continue and I thing that it becomes more and more complicated. Any ideas?
Thank you in advance
Hint. A projective (free) resolution of $k$ as an $R$-module is the following $$\cdots\longrightarrow R^2\stackrel{v}\longrightarrow R^2\stackrel{u}\longrightarrow R^2\stackrel{v}\longrightarrow R^2\stackrel{u}\longrightarrow R^2\stackrel{f}\longrightarrow R\longrightarrow 0$$ where $f(r_1,r_2)=r_1x+r_2y$, $u(a_1,a_2)=(a_1y,a_2x)$, and $v(b_1,b_2)=(b_1x,b_2y)$.