= Tor for a principal non-zero-divisor quotient
{c}
{title2=$\operatorname{Tor}_0^A(A/(a),M)=M/aM,\quad\operatorname{Tor}_1^A(A/(a),M)=\ker(a:M\to M)$}
For a <non-zero-divisor> $a\in A$, resolve $A/(a)$ by $0\to A\xrightarrow{a}A\to A/(a)\to0$. 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 $A/(a)$.
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