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For a connected CW complex $X$ with contractible universal cover, let $\Gamma ={\pi }_1(X)$ and $R=\mathbb{Z}\Gamma$. Lift each cell once and transport its orientation by deck transformations. Its translates form an $R$-basis of the cellular chains of the cover. The augmentation sends every vertex to one, and contractibility makes the augmented chain complex an exact free resolution of the trivial module $\mathbb{Z}$. An $R$-linear cochain with trivial coefficient action is constant on each orbit of lifts, identifying ${\mathrm{Hom}}_R(C_{\ast }(\tilde{X}),\mathbb{Z})$ with the cellular cohomology complex of $X$. Consequently $H^i(X;\mathbb{Z})={\mathrm{Ext}}_R^i(\mathbb{Z},\mathbb{Z})$. The same argument with singular simplex lifts works for spaces with a contractible universal cover.

A double complex has modules $C_{p,q}$ and differentials lowering either index, each squaring to zero and anticommuting with the other. The total chain complex has term ${⨁}_{p+q=i}{C}_{p,q}$ and differential the sum of the two differentials. For the tensor product of two chain complexes, a sign $(-1{)}^p$ on the second differential ensures anticommutation. If the double complex is in the first quadrant and one direction has homology only in degree zero, its total homology is computed by the surviving degree-zero complex. Applying this twice to two free resolutions proves the balanced calculation of the Tor functor.

A coherent sheaf on ${\mathbb{P}}_k^n$ is the sheaf associated with a graded module for a finite graded $k[t_0,\dots ,t_n]$-module. Choose a finite graded free resolution by the Hilbert syzygy theorem and sheafify it; exactness of localization preserves exactness. All terms are finite sums of twisting sheaves on projective space. Twisting sufficiently positively makes every term acyclic in positive degree, and dimension shifting in long exact sequences in sheaf cohomology proves Serre vanishing. The graded-module description is supplied by Stacks Project, Section 30.15.

Start a free resolution of $A/I$ with $P_0=A$ and a free module $P_1$ surjecting onto $I$. The next differential presents the relations in $I$. Right exactness of the tensor product of modules identifies $\mathrm{coker}(P_2{\otimes }_{A}M\to P_1{\otimes }_{A}M)$ with $I{\otimes }_{A}M$. Taking degree-one homology then gives the displayed kernel formula. Hence vanishing for every ideal is equivalent to the flatness criterion using ideals.

Every finitely generated module with a grading over the polynomial ring $S=k[X_1,\dots ,X_n]$ has a finite graded free resolution of length at most $n$. The Koszul resolution of $k$ has length $n$, so ${\mathrm{Tor}}_i^S(M,k)=0$ for $i>n$. In a minimal graded free resolution, this Tor functor is $F_i/(X_1,\dots ,X_n)F_i$; the graded Nakayama lemma forces $F_i=0$ for $i>n$.

Over $S=k[X_1,\dots ,X_n]$ with its standard grading and $\mathfrak{m}=(X_1,\dots ,X_n)$, a graded free resolution is minimal when $d_i(F_i)\subseteq \mathfrak{m}{F}_{i-1}$ for every $i\ge 1$. It is constructed by choosing minimal homogeneous generators for each successive syzygy module, using the graded Nakayama lemma. After tensoring with $k=S/\mathfrak{m}$, all its differentials vanish, so $F_i/\mathfrak{m}{F}_i\cong {\mathrm{Tor}}_i^S(M,k)$; these dimensions and degrees are intrinsic to $M$.

A free resolution of $\mathbb{Q}$ has generators $a_n=1/n!$ and independent relations $a_n-(n+1)a_{n+1}$. Thus the Ext functor gives the cokernel of the map on integer sequence productsThe constant sequence one is not in its image. A preimage would require $u_1\equiv {\sum }_{j=1}^{N}j!(\mathrm{mod}(N+1)!)$ for every $N$. For $N\ge 3$, the sum is below $2N!$, tends to infinity, and its distance from $(N+1)!$ also tends to infinity. No fixed integer satisfies all these congruences. Hence ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})\ne 0$.

Put $\Gamma ={\pi }_1(X,x)$ and $R=\mathbb{Z}[\Gamma ]$. Give $\mathbb{Z}$ its trivial left $R$-module structure, through the augmentation $\epsilon :R\to \mathbb{Z}$, $\sum n_{g}g\mapsto \sum n_g$. The universal cover $\tilde{X}$ has the lifted cell structure, with $\Gamma$ acting by deck transformations.

For each cell of $X$, choose one lifted cell and orient every translate compatibly with its projection. The translates form one copy of the regular $R$-module. Thusis a free left $R$-module, and its boundary is $R$-linear. The augmentation $C_0(\tilde{X})\to \mathbb{Z}$ sends every vertex to one. The contractibility of $\tilde{X}$ says that its reduced homology vanishes, so the augmented complexis exact. It is therefore a free resolution of the trivial module. By the definition of the Ext functor,

An $R$-linear cochain with values in the trivial module has the same value on every translate of a lifted cell. It is therefore exactly an ordinary integral cellular cochain on $X$. The correspondence does not depend on which lift was initially chosen. It also respects the coboundary: applying a constant-on-orbits cochain to the lifted cellular boundary sums precisely the incidence coefficients of the boundary in the base. Henceas cochain complexes, not just as graded groups. Cellular cohomology computes the cohomology of a cell complex, givingAt $i=0$ both sides are $\mathbb{Z}$, by connectedness and ${\mathrm{Hom}}_R(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. This is the cellular free resolution from a contractible universal cover description of group cohomology.

Suppose a nontrivial finite group acts freely on ${\mathbb{R}}^n$ by homeomorphisms. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.

Let $M={\mathbb{R}}^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-manifold with universal cover ${\mathbb{R}}^n$, which is contractible. The group-cohomology argument of Question 2 applies even if no cell structure has been chosen: singular simplices in $M$ lift to the cover, their lifts form free deck-group orbits, and the augmented singular chains of the contractible cover form an exact free resolution. Therefore

We calculate the right-hand side explicitly. Write $R=\mathbb{Z}[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots +t^{p-1}$. The periodic resolution of a finite cyclic group isIts exactness can be seen in coefficients. For $r={\sum }_{j=0}^{p-1}{a}_j{t}^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker (t-1)=\mathbb{Z}N=\mathrm{im}N$. Also $Nr=\epsilon (r)N$, so $\ker N=\ker \epsilon$. If $\epsilon (r)=0$, thenHence $\ker N=\ker \epsilon =(t-1)R$, proving exactness everywhere.

Apply ${\mathrm{Hom}}_R(-,\mathbb{Z})$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The cochain complex is thereforeIts positive even-degree cohomology iswhile its positive odd-degree cohomology is zero.

This contradicts the dimension of the quotient manifold. The noncompact form of Poincaré duality for noncompact manifolds, with the orientation local system when needed, isHere Borel-Moore homology is the homology of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb{Z})=0$ for $i>n$, whether or not $M$ is orientable. Taking any $2j>n$ contradicts the displayed cyclic group calculation. ThusThis proves finite groups cannot act freely on Euclidean space for general homeomorphisms; it does not assume that the action consists of affine maps or isometries.

Use integral coefficients throughout, allowing a nontrivial orientation character. A dimension-$d$ Poincare duality group has a finite resolution by finitely generated projective group-ring modules andas abelian groups. The action on the top copy of $\mathbb{Z}$ may be by signs. This group-ring criterion is particularly suited to the Lyndon–Hochschild–Serre spectral sequence.

Apply that sequence to the extension with coefficient module $\mathbb{Z}G$:As an $N$-module, $\mathbb{Z}G$ is a direct sum of copies of $\mathbb{Z}N$ indexed by the cosets in $Q$. Cohomology of $N$ commutes with this direct sum because its projective resolution is finitely generated. Hence the inner cohomology vanishes except when $b=n$, where it is a direct sum of rank-one groups indexed by $Q$. The $Q$-action shifts these summands freely and transitively, with possible signs from the orientation module. Choosing a generator in one summand and transporting it by $Q$ identifies the resulting left $Q$-module with its regular module $\mathbb{Z}Q$. The residual right group action records the orientation twist; it need not be trivial.

Now the duality of $Q$ says that the only nonzero $E_2$ term is $(a,b)=(q,n)$ and that this term is $\mathbb{Z}$ as an abelian group. No differential can enter or leave it. Therefore $H^{\ast }(G;\mathbb{Z}G)$ has exactly the duality pattern in degree $n+q$. The usual extension-resolution construction also supplies a resolution by finitely generated projective modules: combine the finite kernel and quotient resolutions, lift the quotient differentials and add the necessary homotopy corrections. Its total length is at most $n+q$. Alternatively, the same spectral sequence for arbitrary coefficient modules gives the dimension bound, while the extension finiteness lemma supplies finite generation of the resolution. The surviving top term makes the bound sharp. This proves the extension rule for Poincare duality groups:Requiring a trivial top orientation action would need an additional orientation condition; the untwisted conclusion is not implicit in the hypotheses.

We next need two precise auxiliary dimension results. The Strebel infinite-index subgroup theorem says that an infinite-index subgroup of an integral $P{D}^d$ group has integral cohomological dimension at most $d-1$. The Stallings-Swan theorem says that a group of integral cohomological dimension at most one is free, including the trivial group. Since $G/N$ contains an element of infinite order, $N$ has infinite index in $G$, so $\mathrm{cd}N\le 2$. If its dimension is at most one, the required free-group conclusion follows, with finite rank because $N$ is finitely presented.

Suppose instead that $\mathrm{cd}N=2$. Finite presentation gives a partial free resolution with finitely generated modules through degree two. The kernel in degree one is finitely generated and, because the integral cohomological dimension is two, projective. Truncating at this kernel gives a finite resolution by finitely generated projective modules. Thus $N$ has the finiteness needed for group-ring cohomology to commute with direct sums.

Choose an infinite-order element $\bar{t}\in G/N$, lift it to $t\in G$, and let $H$ be the preimage of $⟨\bar{t}⟩$. Then $H=N⋊⟨t⟩$. Restriction of $\mathbb{Z}H$ to $N$ yields copies of $\mathbb{Z}N$ indexed by all powers of $t$. For every $j$, the quotient generator acts on $H^j(N;\mathbb{Z}H)$ by shifting this direct sum, possibly with an automorphism on each copy. Its invariants vanish, since an invariant finite-support sequence must be zero. Its coinvariants are one copy of $H^j(N;\mathbb{Z}N)$: the shift relations identify successive copies, and any twists can be removed by transporting the identifications along the integer index.

The infinite cyclic group has integral cohomological dimension one, so only columns zero and one occur in the spectral sequence. Column zero vanishes and column one consists of these coinvariants. We obtain the cyclic-extension shift of group-ring cohomologyas abelian groups. Moreover $H^2(N;\mathbb{Z}N)\ne 0$. Here is a useful proof of that last point: if top cohomology of a finite projective resolution were zero, its last dual differential would be onto. Its target is projective, so that differential splits. Dualizing back splits the last injection of the original resolution, shortening it and contradicting integral cohomological dimension two.

It follows that $H^3(H;\mathbb{Z}H)\ne 0$, and hence $\mathrm{cd}H=3$. Strebel's theorem forces $H$ to have finite index in $G$. It is therefore itself a $P{D}^3$ group, by finite-index invariance of Poincare duality for torsion-free groups. The displayed shift now makes $H^j(N;\mathbb{Z}N)$ zero except for $j=2$, where it is $\mathbb{Z}$. Together with its finite projective resolution, this provesThe classification of two-dimensional Poincare duality groups identifies the latter with the fundamental group of a closed aspherical surface. This classification includes orientation-twisted surface groups; it does not include the two-sphere or real projective plane, whose fundamental groups do not have integral cohomological dimension two.

In this latter case we have also proved that $G/N$ is virtually infinite cyclic, since $H/N=⟨\bar{t}⟩$ has finite index. If necessary replace the surface group by its characteristic index-two orientable subgroup and take the subgroup generated by it and $t$. This still has finite index in $G$ and is a semidirect product of an orientable closed aspherical surface group by $\mathbb{Z}$. The Dehn-Nielsen-Baer theorem for closed orientable surfaces realizes the outer automorphism defined by conjugation by $t$ as a surface homeomorphism $f$. For the torus this is just realization of $G{L}_2(\mathbb{Z})$ by linear homeomorphisms. Changing the chosen lift accounts for an inner automorphism. The mapping torusis a surface bundle over $S^1$ with fundamental group ${\pi }_1(S){⋊}_{f_{\ast }}\mathbb{Z}$. Thus a finite-index subgroup of $G$ is the fundamental group of a surface fibration over the circle. Taking a further double cover of the base if required makes the monodromy orientation-preserving.

Here the usual homological algebra convention is that the right-exact functor $F$ is additive. Module categories have enough projective modules, because any module is a quotient of a free module. Choose a projective resolutionApply $F$ to the unaugmented chain complex and define the left derived functors byAdditivity ensures that $F$ preserves zero maps, sums of maps and homotopy identities, so the transformed differentials still compose to zero.

For a module homomorphism $u:M\to N$, lift it to a chain map between chosen projective resolutions. In degree zero, lift through $Q_0\twoheadrightarrow N$ using projectivity of $P_0$. In each higher degree, the already constructed map lands in the next kernel; exactness of $Q_{\bullet }$ and projectivity of $P_i$ supply the next lift. Any two lifts are chain homotopic: their degree-zero difference lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1^Q$, so it lifts to $h_0:P_0\to Q_1$. Inductively lift the remaining difference after subtracting the preceding homotopy term. This givesApplying $F$ preserves this identity, so the lifts induce the same homology map. Identity lifts and composable lifts show that $L_{i}F$ is a functor. Comparison lifts between two resolutions over the identity of $M$ are inverse up to chain homotopy; hence the construction is independent of the chosen resolution up to its canonical natural isomorphism.

Right exactness identifies $L_{0}F(M)=\mathrm{coker}(F(P_1)\to F(P_0))$ naturally with $F(M)$. For a projective module $P$, the length-zero resolution gives $L_{i}F(P)=0$ for $i>0$. For a short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, the horseshoe lemma builds a degreewise split short exact sequence of projective resolutions. Concretely, lift the projective generators covering $M^{\prime \prime }$ into $M$ and combine them with generators covering $M^{\prime }$; repeat on the kernels. The middle projective term in each degree is the direct sum of the two outer ones. An additive functor preserves these split sequences, and lifting cycles, taking their differentials and returning to the subcomplex produces the connecting maps inThis is the natural long exact derived-functor sequence. In particular, if $0\to K\to P\to M\to 0$ has $P$ projective, then $L_{i}F(M)\cong L_{i-1}F(K)$ for $i\ge 2$, and $L_{1}F(M)=\ker (F(K)\to F(P))$. These are dimension-shifting formulas. Positive degrees can be killed by the surjection $P\to M$. If $F$ is exact, it preserves every resolution's exactness and all its positive left derived functors vanish. Conversely, if $L_{1}F$ vanishes on all modules, the displayed long exact sequence makes $F$ exact. A natural transformation between additive right-exact functors induces transformations on the left derived functors by applying it termwise.

For extension of scalars $F(M)=B{\otimes }_{A}M$, these derived functors are the Tor functors ${\mathrm{Tor}}_i^A(B,M)$. When $B=A/(a)$ and $a$ is a non-zero-divisor, there is the length-one free resolutionAlthough the construction above resolves $M$, one may instead use this resolution of $B$. To justify that calculation, tensor a free resolution of $M$ with the two-term resolution of $B$ to form a double complex. Taking homology first along the resolution of $B$ leaves $B{\otimes }_A{P}_{\bullet }$; taking it first along the resolution of $M$ leaves $M\stackrel{a}{\to }M$. The free terms preserve exactness under tensoring. The double complex is first-quadrant with finitely many terms on each total diagonal, so both filtrations compute the same total homology. This proves the required balancing directly.

The kernel and cokernel of multiplication by $a$ now give Tor for a principal non-zero-divisor quotient:These are naturally $B$-modules. For $M=A$, the degree-one group is zero; for $M=B$, multiplication by $a$ is zero and both degrees zero and one are $B$. Thus, when $B\ne 0$, degree one explicitly detects the failure of this base-change functor to be exact.

For a left $\mathbb{Z}G$-module, group cohomology iswith $\mathbb{Z}$ the trivial module. Equivalently take a projective resolution $P_{\bullet }\to \mathbb{Z}$ and the cohomology of ${\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },M)$. This also describes the right-derived functors of the invariant-module functor $M\mapsto M^G$.

For an explicit model, the homogeneous bar free resolution has $P_n=\mathbb{Z}[G^{n+1}]$, with diagonal $G$-action and boundary the alternating sum of coordinate deletions. It is free over $\mathbb{Z}G$: each orbit has a unique representative with first coordinate $1$. The augmented complex is exact because inserting $1$ as a first coordinate supplies a contracting homotopy of underlying abelian groups; its lack of $G$-equivariance does not affect exactness. The associated inhomogeneous group cochains are functions $f:G^n\to M$, withFor group cochains of degree zero, $(dm)(g)=gm-m$. Adjacent deletions cancel in pairs, giving $d^2=0$ and hence $H^n=\ker d/\mathrm{im}d$.

In degree zero this givesIt is also ${\mathrm{Hom}}_{\mathbb{Z}G}(\mathbb{Z},M)$, since a map is determined by the invariant image of $1$.

In degree one the cocycle condition is $f(gh)=f(g)+gf(h)$, soThese are crossed homomorphisms modulo principal crossed homomorphisms. For trivial action they reduce to ordinary group homomorphisms, so $H^1(G,M)=\mathrm{Hom}(G_{\mathrm{ab}},M)$.

There are two further useful descriptions. If $I=\ker (\mathbb{Z}G\to \mathbb{Z})$ is the augmentation ideal, a crossed homomorphism gives an equivariant map $I\to M$ by $g-1\mapsto f(g)$, because $gh-1=(g-1)+g(h-1)$. Applying $\mathrm{Hom}(-,M)$ to $0\to I\to \mathbb{Z}G\to \mathbb{Z}\to 0$ therefore givesAlso a section of $M⋊G\to G$ has the form $g\mapsto (f(g),g)$ and is a group homomorphism exactly when $f$ is a crossed homomorphism. Conjugation of sections by an element of $M$ changes $f$ by a principal crossed homomorphism. Thus $H^1$ classifies the $M$-conjugacy classes of such sections, or equivalently of complements to $M$ in the semidirect product.

In degree two, a normalized two-cocycle $c:G\times G\to M$ satisfiesDefine multiplication on $M\times G$ byThe displayed cocycle identity is precisely associativity; normalization gives the identity and inverses exist. This produces a group extension $1\to M\to E_c\to G\to 1$ with the prescribed action on $M$.

Conversely a set-theoretic section $s:G\to E$, $s(1)=1$, gives $c(g,h)=s(g)s(h)s(gh{)}^{-1}$ in the additive kernel of a group homomorphism. Associativity gives the cocycle condition. Replacing $s(g)$ by $b(g)s(g)$ changes $c$ byThe resulting extensions are equivalent through a change of coordinates in $M\times G$. Conversely any extension equivalence preserving both kernel of a group homomorphism and quotient yields such a change of section. HenceThe zero class corresponds exactly to a split extension. With trivial action these are central group extensions; addition of classes corresponds to the Baer sum. This proves the second group cohomology classifies group extensions interpretation.

For $G={\mathbb{Z}}^2$ and trivial coefficients $\mathbb{Z}$, put $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The Koszul resolution for a rank-two free abelian group iswhere $d_1(a,b)=(x-1)a+(y-1)b$ and $d_2(c)=(-(y-1)c,(x-1)c)$. To check exactness at $R^2$, reduce $d_1(a,b)=0$ modulo $x-1$. Since $y-1$ is not a zero divisor in $\mathbb{Z}[y^{\pm 1}]$, write $b=(x-1)c$, and cancellation gives $a=-(y-1)c$. Injectivity of $d_2$ follows because $R$ is a domain. The augmentation kernel is generated by $x-1,y-1$, proving the other exactness statement.

Applying ${\mathrm{Hom}}_R(-,\mathbb{Z})$ makes both differentials zero, since $x,y$ act trivially. ThusDegree-one classes assign arbitrary integers to the two generators. Geometrically these are also the cohomology groups of the torus, a classifying space for ${\mathbb{Z}}^2$; its cohomology ring is the exterior algebra on the two degree-one classes.

The integer $n$ in degree two has an explicit extension representative. On triples $(a,b,m)\in {\mathbb{Z}}^3$, setThe bilinear term is a two-cocycle. With generators $x=(1,0,0)$, $y=(0,1,0)$ and central $z=(0,0,1)$,Every central extension of ${\mathbb{Z}}^2$ by $\mathbb{Z}$ has this form: choose lifts of the two generators and the fixed generator of the kernel of a group homomorphism, then their group commutator determines $n$ and every element has a unique form $z^m{y}^b{x}^a$. Changing the lifts by central elements leaves that commutator unchanged. Thus distinct integers give inequivalent extensions, and the cocycles $nad$ add with $n$. The case $n=0$ is ${\mathbb{Z}}^3$; $n=1$ is the integer Heisenberg group. This realizes the integral central extensions of a rank-two free abelian group concretely.

To compute the first Tor of a cyclic quotient, start a free resolution of $A/I$ with $P_0=A$, take a free module $P_1$ surjecting onto $I$, and take $P_2$ surjecting onto the kernel of $P_1\to I$. Tensor right exactness identifiesBy definition, the degree-one Tor functor is the kernel of $P_1{\otimes }_{A}M\to A{\otimes }_{A}M$ modulo the image from $P_2{\otimes }_{A}M$. Quotienting first by that image consequently gives the natural identificationEquivalently the calculation gives the exact sequenceThus its first term vanishes for every ideal exactly when each multiplication map is injective, provingTogether with the preceding implications, all four conditions are equivalent. The proof used explicit extension arguments, character duality and the definition of the Tor functor, rather than an unproved flatness or injectivity criterion.

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 a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.

The direct image of a coherent sheaf under a closed immersion puts $\mathcal{G}=f_{\ast }\mathcal{F}$ in the category of coherent sheaves on ${\mathbb{P}}_k^n$. By sheaf cohomology under a closed inclusion and the supplied compatibility of twisting with direct image,Here is a proof of the required Serre vanishing on projective space. A coherent sheaf on ${\mathbb{P}}_k^n$ is the sheaf associated with a graded module for a finite graded $k[t_0,\dots ,t_n]$-module. Equivalently, it has a presentation by finite sums of twisting sheaves. Use a finite twisting resolution of a coherent sheaf on projective space: resolve the graded module by a finite graded free resolution, using the Hilbert syzygy theorem, and sheafify; exactness of localization preserves the resolution. Its terms are finite sums of $\mathcal{O}(a)$.

Choose $d$ sufficiently large that all twists $a+d$ occurring in these finitely many terms are nonnegative. The cohomology of twisting sheaves on projective space then vanishes in every positive degree for every resolution term. In a short exact sequence $0\to {\mathcal{K}}_{j+1}\to {\mathcal{E}}_j\to {\mathcal{K}}_j\to 0$, with ${\mathcal{K}}_0=\mathcal{G}$, the long exact sequence in sheaf cohomology identifies $H^p({\mathcal{K}}_j(d))$ with $H^{p+1}({\mathcal{K}}_{j+1}(d))$ for $p>0$. Iterating to the final acyclic term provesThe standard graded-module description used here is given in Stacks Project, Section 30.15; the vanishing follows from the displayed finite-resolution argument.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that