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 $\{xe_1 , x^2e_2,....\}$.

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


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