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A short exact sequence presenting a module as a quotient of a projective module. The kernel need not be projective. Continuing a projective presentation gives a projective resolution.

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 short exact sequence $0\to U\stackrel{i}{\to }V\stackrel{q}{\to }W\to 0$ and any module $E$, the sequence $U{\otimes }_{A}E\to V{\otimes }_{A}E\to W{\otimes }_{A}E\to 0$ is exact. The last map is surjective since every pure tensor $w\otimes e$ can be lifted using a preimage of $w$. Put $T=(V{\otimes }_{A}E)/\mathrm{im}(i\otimes 1_E)$. There is a well-defined balanced map $W\times E\to T$, sending $(q(v),e)$ to the class of $v\otimes e$: replacing $v$ by $v+i(u)$ changes it by an element of the discarded image. This induces $W{\otimes }_{A}E\to T$, inverse to the map induced by $q\otimes 1_E$. Thus the kernel of $q\otimes 1_E$ is exactly the image of $i\otimes 1_E$. Consequently a module is flat exactly when tensoring with it also preserves injections.

A degreewise short exact sequence of cochain complexes induces a long exact sequence from a coefficient sequence in the coefficient case, and a long exact cohomology sequence generally. Its degree-one connecting homomorphism lifts a quotient cocycle, takes its coboundary, and identifies that result in the subcomplex.