# Finding Tor of k[x]-module

I am asked to find $$\operatorname{Tor}_{*}^{k[x]}(M,M)$$, with $$M=k[x,x^{-1}]/xk[x]$$.

I start with finding a projective resolution for $$M$$. An arbitrary element of $$M$$ is $$\sum_{n \leq 0}a_nX^{-n}$$, so I was looking at a surjection $$\oplus_{i=1}^{\infty} k[x] \to M$$, where we map $$(1,0,0,\ldots) \mapsto 1$$ $$(0,1,0,0,\ldots) \mapsto x^{-1}$$ $$(0,0,1,0,0,\ldots) \mapsto x^{-2}$$ and more generally $$e_i \mapsto x^{-i+1}$$ and extend this linearly.

The kernel of this map is generated by elements of the form $$x(1,0,0,\ldots) ,x^2(0,1,\ldots),\ldots , x^{i}e_i$$.

We continue the projective resolution by finding a map surjecting on this kernel, ie $$\oplus_{i=1}^{\infty} k[x] \to \oplus_{i=1}^{\infty}k[x]$$, where we define $$(1,0,0,\ldots) \mapsto x(1,0,0,\ldots)$$ $$(0,1,0,\ldots) \mapsto x^2(0,1,0,\ldots)$$ and more generally $$e_i \mapsto x^{i}e_i$$ and then extend this linearly. We thus have $$P_{*} : 0 \to \oplus_{i=1}^{\infty} k[x] \to \oplus_{i=1}^{\infty} k[x] \to 0$$.

We can then tensor this with $$M$$, so $$P_{*}$$ becomes $$0 \to \oplus_{i=1}^{\infty} M \to \oplus_{i=1}^{\infty} M \to 0$$ with the induced map being the zero map. So I guess this gives $$\operatorname{Tor}_{*}^{k[x]}(M,M)$$.

Something doesn't seem right to me, but I am not sure I can find where this argument is flawed. Any ideas?

EDIT: I guess I have found the first error in this argument. It's the kernel of the first map $$\oplus_{i=1}^{\infty} k[x] \to M$$ , i.e. $$e_1 -xe_2$$ is in the kernel, but that is not generated by $$\{xe_1 , x^2e_2,....\}$$.

• Indeed, the kernel of your map $\bigoplus k[x]\to M$ is rather generated by $e_i-xe_{i+1}$ for all $i$, and $xe_1$. Jan 18, 2020 at 13:43
• Have you tried using the exact sequence $0\to k[x]\xrightarrow{\cdot x} k[x,x^{-1}]\to M\to 0$? Jan 18, 2020 at 13:51
• It definitely crossed my mind, though I wasn't sure how to prove that $k[x,x^{-1}]$ is projective, if at all. Jan 18, 2020 at 13:55
• It is not projective (otherwise it would be free), but it is flat, which is enough. Jan 18, 2020 at 14:11