Let $R$ be a commutative ring with unity. Let $M$ be a finitely generated $R[t]$-module with a projective resolution $$ 0 \longrightarrow P \longrightarrow Q \longrightarrow M \longrightarrow 0 $$ where $P$, $Q$ are finitely generated projective $R[t]$-modules.
Now it is given that $M$ is killed by some power of $t$, let's say $t^n$ kills $M$.
To show that $\mathrm{Tor}^{R[t]}_{1}(M,R[t]/(t^n))$ is $M$ itself.
My attempt:
Given a short exact sequence I always have a long exact sequence of $\mathrm{Tor}$, thus I have $$0\longrightarrow \mathrm{Tor}^{R[t]}_{1}(M,R[t]/(t^n)) \longrightarrow P \otimes R[t]/(t^n) \longrightarrow Q \otimes R[t]/(t^n) \longrightarrow M \otimes R[t]/(t^n) \longrightarrow 0.$$
Now for any $R[t]$-module $N$ we have $N \otimes R[t]/(t^n) \cong N/t^nN$.
So I know that $\mathrm{Tor}^{R[t]}_{1}(M,R[t]/(t^n))$ is nothing but the kernel of $$P/t^nP \longrightarrow Q/t^nQ.$$
Here I have some trouble in calculating the kernel and showing it is isomorphic to $M$.
I think the isomorphism can be shown from here as well but it seems I am missing something very obvious
$$0\longrightarrow \mathrm{Tor}^{R[t]}_{1}(M,R[t]/(t^n)) \longrightarrow P/t^nP \longrightarrow P /t^nQ \longrightarrow 0$$
For any help or hint I am grateful.
Best Answer
One has the exact sequence $$0\to t^n R[t]\to R[t]\to R[t]/t^n R[t]\to0.$$ Tensoring with $M$ gives a long exact sequence $$\cdots\to\text{Tor}_1^{R[t]}(M,R[t]) \to\text{Tor}_1^{R[t]}(M,R[t]/t^n R[t])\to M\otimes_{R[t]}t^n R[t[\to M\otimes_{R[t]}R[t]\to\cdots. $$ Since $R[t]$ is projective, $\text{Tor}_1^{R[t]}(M,R[t])=0$ and this reduces to $$\cdots\to0 \to\text{Tor}_1^{R[t]}(M,R[t]/t^n R[t])\to M\stackrel{\times t^n}\longrightarrow M\to\cdots$$ which gives $\text{Tor}_1^{R[t]}(M,R[t]/t^n R[t])$ as the $t^n$-torsion of $M$.
Under your hypotheses, that is all of $M$.