Tor for a principal non-zero-divisor quotient (${\mathrm{Tor}}_0^A(A/(a),M)=M/aM,{\mathrm{Tor}}_1^A(A/(a),M)=\ker (a:M\to M)$)

For a non-zero-divisor $a\in A$, resolve $A/(a)$ by $0\to A\stackrel{a}{\to }A\to A/(a)\to 0$. Tensoring with $M$ gives the kernel and cokernel in the displayed formula; all higher Tor functors vanish. Resolving either argument gives the same homology by the two filtrations of the double complex of the two free resolutions. For $M=A/(a)$, both degrees zero and one equal $A/(a)$.