I have a question about the following portion of the proof of a result
The proof is given by a serious of reductions, but I got stuck at the first step, where it claims that if $x\in R$ is a non zero-divisor on $M$ (and $R$), then ${\rm Tor}^R_i(M, N)\simeq{\rm Tor}^{R/x}_i(M/x, N/x)$.
I guess we need to take a free resolution $L_{\bullet}$ of $M$, and then apply $\otimes_RR/x$, but I can't confirm that $L_{\bullet}/x$ a (free) resolution of $M/x$.
Exactness at the end is just by right exactness of tensor product. So we still need to prove something like: If $L'\to L\to L''$ is an exact sequence of free $R$-modules, then $L'/x\to L/x\to L''/x$ is also exact. I can't prove this, and suspect if it is true.
Maybe the result should be proven by some other method. Some reference of proof of this result is also fine. Thanks for any help!
$\newcommand{\Tor}{\operatorname{Tor}}$One can make the comment above into an answer: pick a free resolution $P_*$ of $M$ and one $Q_*$ of $N$ and consider the double complex $C_{*,*} = P_* \otimes_R Q_* = (P_* \otimes_R R/x)\otimes_{R/x} Q_*$.
We can filter it in such a way that the homology of the associated graded is equal to $H_*(P_* \otimes_R R/x)\otimes_{R/x} Q_*$. The first factor computes $\Tor_*^R(M,R/x)$, and since $x$ is a non-zero divisor in $R$ we can use the resolution $R\longrightarrow R$.
This shows that $\Tor_0(M,R/x) = M/xM$ and $\Tor_1(M,R/x) = M_x$ the $x$-torsion part of $M$, which is zero, and all other groups vanish. Thus the homology of the associated graded is the (no longer double) complex $M/xM\otimes_{R/x} Q_*$ whose homology is of course $\Tor_*^{R/x}(M/xM,N)$.