= Ext from a principal non-zero-divisor quotient
{c}
{title2=$\operatorname{Ext}_A^0(A/(a),M)=\ker a,\quad\operatorname{Ext}_A^1(A/(a),M)=M/aM$}
For a non-zero-divisor $a\in A$, the free resolution $0\to A\xrightarrow{a}A\to A/(a)\to0$ gives degree-zero kernel and degree-one cokernel, and all higher <Ext functors> vanish. Directly in an <injective resolution> of the second variable, multiplication by $a$ is surjective in every term: the map $aA\to I$ taking $ax$ to $xm$ extends to $A\to I$. Thus $0\to\operatorname{Hom}_A(A/(a),I^\bullet)\to I^\bullet\xrightarrow{a}I^\bullet\to0$ is a short exact sequence of complexes and gives the same result. The displayed Hom functor is covariant in $M$; changing it to $\operatorname{Hom}_A(M,A/(a))$ changes the variable that is resolved.
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