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A regular sequence in a commutative ring $A$ is a finite list such that $(f_1,\dots ,f_r)$ is proper and $f_j$ is a non-zero-divisor on $A/(f_1,\dots ,f_{j-1})$ for each $j$. In a graded ring with homogeneous $f_j$, injectivity can be checked on homogeneous elements because components of distinct degrees cannot cancel. It gives short exact sequences $0\to R_{j-1}(-\mathrm{deg}{f}_j)\to R_{j-1}\to R_j\to 0$ for the successive graded quotients.

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 $A/(a)$.

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.