Let $k$ be a field and let $R = k[x]$. Consider the $R$-module $M := \frac{k[x, x^{-1}]}{x \cdot k[x]}$ (i.e. so typical elements are Laurent polynomials with no positive powers).
I have computed $\text{Tor}_1^R(M, M) = M$ (see below); is this correct? Is the proof below correct?
Let $L$ denote the $R$-module $k[x, x^{-1}]$. Then we have a short exact sequence
$$0 \longrightarrow R \xrightarrow{1\, \mapsto \,x} L \longrightarrow M \longrightarrow 0$$
This induces the long exact sequence
$$ \begin{alignat}{3} \cdots & \longrightarrow \text{Tor}_1^R (M, L) && \longrightarrow \text{Tor}_1^M(M, M) & & \longrightarrow \\ M \otimes_R R & \longrightarrow M \otimes_R L && \longrightarrow M \otimes_R M & & \longrightarrow 0\\ \end{alignat} $$
Since $L$ is flat, $\text{Tor}_1^R (M, L) = 0$.
Also, consider an element $(x^{-n} \otimes_R x^{-m}) \in M \otimes_R M$. We have
$$ x^{-n} \otimes x^{-m} = x^{-n} \otimes (x^{n+1} \cdot x^{-n - m - 1}) = (x^{n+1} \cdot x^{-n}) \otimes x^{-n -m -1} = 0 $$
so that $M \otimes_R M = 0$. Similarly, $M \otimes_R L = 0$.
Hence, the long exact sequence reduces to
$$0 \longrightarrow \text{Tor}_1^R(M, M) \longrightarrow M \longrightarrow 0$$
That is, $\text{Tor}_1^R(M, M) \simeq M$.