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Let $(R,\mathfrak{m})$ be a Noetherian local ring, let $M$ be nonzero and finitely generated, and let every $f_i$ lie in $\mathfrak{m}$. Then positive Koszul homology vanishes exactly when $f$ is a regular sequence on a module $M$. The mapping cone exact sequence proves the forward implication by induction; surjectivity of the last generator on earlier homology and the Nakayama lemma prove the converse.

Tensor the finite free Koszul complex on $f_1,\dots ,f_r$ with the module $M$. For a regular sequence on a module, its positive Koszul homology vanishes. If the sequence is regular on $R$, this complex instead computes ${\mathrm{Tor}}^R(R/(f),M)$ for arbitrary $M$.

Let $R$ be a commutative ring and $f=(f_1,\dots ,f_r)$. The Koszul complex packages these elements and their relations in a finite chain complex of free modules. With $E=R^r$ and basis $e_1,\dots ,e_r$, set $K_i(f;R)={\bigwedge }^{i}E$ for $0\le i\le r$, and defineDeleting two distinct basis elements in the two possible orders gives opposite signs and the same coefficient, so ${\partial }^2=0$. Equivalently $K(f;R)$ is the tensor product of chain complexes of $r$ two-term complexes $[R\stackrel{f_i}{\to }R]$ in degrees one and zero. The tensor differential is $\partial (u\otimes v)=\partial u\otimes v+(-1{)}^{\mathrm{deg}u}u\otimes \partial v$. This fixes the sign convention. On the exterior algebra, the differential is a graded derivation determined by $\partial e_i=f_i$.

For an $R$-module $M$, define the Koszul complex with module coefficients by $K(f;M)=K(f;R){\otimes }_{R}M$. Its degree-zero Koszul homology isIts higher Koszul homology measures the failure of these equations to form a regular sequence on a module. Each $f_j$ annihilates every homology group. Indeed the Koszul homotopy for multiplication by a generator is $h_j(u)=e_j\wedge u$, and the graded product rule givesFor coefficients in $M$ the same identity holds after tensoring. Thus multiplication by $f_j$ is zero on homology, and the homology groups are naturally modules over $R/(f)$. If $(f)=R$, choose ${\sum }_j{a}_j{f}_j=1$; then ${\sum }_j{a}_j{h}_j$ is a contracting homotopy, so the entire complex is contractible. These identities are also useful after localization of a ring: wherever one generator is a unit, the complex has zero homology.

The construction is functorial under a ring homomorphism, and base change gives $K(f;R){\otimes }_{R}S=K(\bar{f};S)$, since its terms are free with the displayed basis and differential. An invertible change of generators $f_i={\sum }_j{u}_{ij}{g}_j$ gives a chain isomorphism by sending $e_i$ to ${\sum }_j{u}_{ij}{e}_j$ and extending to exterior powers. In particular, permuting the generators changes only the exterior signs, not the isomorphism class of the complex.

The essential exactness theorem is that a regular sequence on a module gives zero positive Koszul homology. Recall that $f_i$ must act injectively on $M/(f_1,\dots ,f_{i-1})M$, with final quotient nonzero. Put $f^{\prime }=(f_1,\dots ,f_{r-1})$. Adding the final two-term complex identifies $K(f;M)$ with the mapping cone of multiplication by $f_r$ on $K(f^{\prime };M)$, with the chosen tensor signs. The long exact sequence in homology consequently has segmentsFor $r=1$, positive homology is exactly $\ker (f_1:M\to M)$. By induction the preceding homology vanishes above zero. The final injectivity assumption kills $H_1(f;M)$, while the exact sequence kills every $H_i$ for $i>1$. Thereforefor a regular sequence on a module. In particular $K(f;R)$ is a finite free resolution of $R/(f)$ when $f$ is a regular sequence, with ranks $\left(\genfrac{}{}{0ex}{}{r}{i}\right)$ and length $r$.

There is a precise converse under local finiteness assumptions. Suppose $(R,\mathfrak{m})$ is a Noetherian local ring, $0\ne M$ is finitely generated, and all $f_i\in \mathfrak{m}$. If positive Koszul homology vanishes, the same exact sequence makes multiplication by $f_r$ surjective on every $H_i(f^{\prime };M)$ for $i>0$. These modules are finitely generated because the ring is Noetherian and the terms of the complex are finite modules. The Nakayama lemma gives $H_i(f^{\prime };M)=0$. Induction makes $f^{\prime }$ regular on $M$, and the degree-one part of the exact sequence makes $f_r$ injective on $H_0(f^{\prime };M)$. The final quotient is nonzero, again by the Nakayama lemma. This proves the Koszul acyclicity criterion in a Noetherian local ring:The hypotheses matter: a unit ideal makes the complex contractible but cannot be a proper regular sequence.

Examples make both sides visible. For $R=k[x,y]$ and $f=(x,y)$, the Koszul resolution isThe signs agree with $\partial (e_1\wedge e_2)=x{e}_2-y{e}_1$. For $R=k[x]$ and $f=(x,x)$, the degree-one cycles are $R(1,-1)$ and the boundaries are $xR(1,-1)$; hence $H_1\cong R/(x)$, while $H_2=0$. The repeated element has exposed a relation that the first element already kills.

The Koszul complex also computes derived functors. If $f$ is a regular sequence on $R$, its free resolution givesHere no regularity of $f$ on $M$ is assumed. For $S=k[x_1,\dots ,x_r]$, tensor the Koszul resolution on the variables with $k$. Every differential becomes zero, soThe top group is nonzero, showing that a free resolution of $k$ cannot be shorter than $r$.

Finally, wedging complementary exterior degrees gives a perfect pairing $K_i{\otimes }_R{K}_{r-i}\to {\bigwedge }^r{R}^r\cong R$. It identifies the dual cochain complex with the degree-reversed Koszul complex, after the appropriate signs. This self-duality of the Koszul complex shows, for a regular sequence, thatThus the same explicit complex simultaneously records quotient equations, regularity, relations, Tor functor computations and Ext functor computations. Its finiteness and exterior structure are what make it especially effective in commutative algebra.

For central $t_1,\dots ,t_n$, define the Koszul complex on central ring elements using formal exterior symbols:with zero terms outside $0\le k\le n$ and differentialThe terms are free bimodules with the formal symbols commuting with coefficients. Centrality makes $d$ a bimodule map, and the terms in $d^2$ cancel in pairs because the $t_i$ commute. This defines the Koszul complex over a possibly noncommutative ring without using an undefined exterior algebra of arbitrary one-sided modules. In particular $K(\mathbf{t}){\otimes }_{R}M$ is a well-defined chain complex of left modules.

A regular sequence on a module means that multiplication by $t_i$ is injective on $M/(t_1,\dots ,t_{i-1})M$ for each $i$, usually with the additional convention that the final quotient is nonzero. The homology vanishing below uses only the injectivity conditions. For one element, the chain complex is $M\stackrel{t_1}{\to }M$, with zero first homology and zeroth homology $M/t_{1}M$.

For the induction let $C=K(t_1,\dots ,t_{n-1}){\otimes }_{R}M$. Adjoining the last generator identifies the new Koszul complex with the mapping cone of multiplication by $t_n$ on $C$. In the convention $\mathrm{Cone}(t_n{)}_k=C_k\oplus C_{k-1}$ its differential is $(a,b)\mapsto (da+t_{n}b,-db)$; the identification is $a+e_n\wedge b$. The long exact sequence in homology of this mapping cone givesThe induction hypothesis is $H_k(C)=0$ for $k>0$ and $H_0(C)=N=M/(t_1,\dots ,t_{n-1})M$. Regularity makes $t_n:N\to N$ injective. Hence the new positive homology is zero and its zeroth homology is $N/t_{n}N$. The augmentation to this quotient induces these homology isomorphisms, giving the quasi-isomorphismwhere the right side is placed in degree zero. The results used are the long exact sequence in homology of a degreewise short exact sequence of complexes and its mapping cone form, together with the stated induction; no unmentioned acyclicity criterion is needed.

For $R=\mathbb{Z}[X]$, $(p,X)$ is a regular sequence: $p$ is a non-zero-divisor in $R$, and multiplication by $X$ is injective in $(\mathbb{Z}/p\mathbb{Z})[X]$, even when $p$ is composite. The Koszul resolution of $R/(p,X)$ isWith $N=\mathbb{Z}/q\mathbb{Z}$ and $X$ acting as zero, applying the Hom functor gives the cochain complexThus, more generally, the Ext groups between polynomial-ring residue modules areand vanish in all degrees above two. For the equal-modulus case,These are $R$-modules through $X=0$ and reduction modulo $p$. If the multiplicative self-Ext functor is desired, the Koszul self-Ext algebra is the exterior algebra on two degree-one generators over $\mathbb{Z}/p\mathbb{Z}$. The two contractions on the Koszul resolution lift these classes, square to zero and anticommute, so this description also holds for composite $p$ and characteristic two. For coprime $p,q$, multiplication by $p$ is invertible on $N$, so both its kernel and cokernel vanish: