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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.

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 $i>n$.

The cellular cochain complex of finite Real projective space consists of alternating maps zero and two. Its groups can be tensored using the Künneth theorem; the integral Künneth torsion polynomial efficiently counts all tensor and Tor functor summands. A product of factors of dimensions two, three and four has free-rank polynomial $1+t^3$ and torsion polynomial $3t^2+3t^3+5t^4+6t^5+5t^6+4t^7+2t^8+t^9$.

Each real projective plane has cells $e_0,e_1,e_2$ with $d{e}_2=2e_1$ and $d{e}_1=0$. Their product cellular chain complex has ranks $1,2,3,2,1$. Its degree-two cycle $e_1\times e_1$ has order two in homology. The degree-three cycle $e_2\times e_1+e_1\times e_2$ also has order two, because its double is $d(e_2\times e_2)$. Consequently $H_0=\mathbb{Z}$, $H_1=(\mathbb{Z}/2{)}^2$, $H_2=H_3=\mathbb{Z}/2$ and all other groups vanish. The degree-three torsion is also the Tor functor term in the Künneth theorem.

For spaces whose integral cohomology has only free and $\mathbb{Z}/2$ summands, let $F$ encode free ranks and $T$ count order-two summands by degree. An additively split integral Künneth theorem gives $F_{A\times B}=F_A{F}_B$ and the displayed rule for $T$. The $t^{-1}$ term is the cohomological degree shift of the Tor functor contribution. The formula requires this torsion type and the usual finite-type hypotheses.

For an additive right-exact functor on module categories, apply it to a projective resolution and take homology. Lifts of module maps exist by projectivity; any two lifts are chain homotopic, giving independence and functoriality. The zeroth derived functor is naturally $F$, positive degrees vanish on projective modules, and the horseshoe lemma produces a long exact homology sequence from each short exact sequence. A Tor functor is the left derived functor of a tensor product of modules functor.

Use the standard CW complex structure on the real projective plane, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the cellular chain complex has $d{e}_2=2e_1$ and $d{e}_1=0$.

The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le 2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The cellular chains of a product of finite CW complexes have boundary $d(e_i\times e_j)=d{e}_i\times e_j+(-1{)}^i{e}_i\times d{e}_j$. In the ordered basesthe nonzero matrices areIn particular $d_2{d}_3=d_3{d}_4=0$. The degree-one homology is ${\mathbb{Z}}^2/2{\mathbb{Z}}^2$. The degree-two cellular cycles are $\mathbb{Z}{e}_{11}$, and the cellular boundaries there are $2\mathbb{Z}{e}_{11}$. The degree-three cycles are $\mathbb{Z}(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the integral homology of two real projective planes:The degree-three torsion is the Tor functor contribution in the Künneth theorem; it would be missed by retaining only tensor products of the two factors' homology groups.

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.

Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.

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.

Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.

For a cohomological spectral sequence, $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ and the next page is its cohomology. In the bounded setting, convergence of a spectral sequence to $H^{\ast }$ means that each $H^n$ has a finite decreasing, exhaustive and separated filtration withHere $E_{\infty }^{p,q}$ is the eventual stable value. This determines the associated graded module of $H^n$, rather than automatically a canonical direct-sum decomposition of $H^n$. For unbounded filtrations additional completeness/convergence conditions are necessary; boundedness removes those issues here.

The bounded filtered-complex convergence theorem states the following. If $F^{p}C$ is a decreasing filtration by cochain subcomplexes, preserved by the differential, finite in each cochain degree (uniform bounds $F^{a}C=C$, $F^{b}C=0$ are sufficient), then there is a spectral sequenceThe abutment filtration is $F^p{H}^n(C)=\mathrm{im}[H^n(F^{p}C)\to H^n(C)]$. Its finite length ensures stabilization and the displayed limiting-page identification. The filtered cochain complex need not itself be bounded in cochain degree. In the degreewise finite version, the bounds in degrees $n-1,n,n+1$ suffice to stabilize the terms of total degree $n$.

For a double cochain complex $D^{p,q}$ bounded in both indices, form the total cochain complex ${\mathrm{Tot}}^{n}D={⨁}_{p+q=n}{D}^{p,q}$ with anticommuting differentials $d_h,d_v$ and total differential $d_h+d_v$. If the original differentials commute, inserting the usual sign in one of them gives this convention. Filtering by the first index gives $E_1^{p,q}=H_v^q(D^{p,\ast })$ and then $E_2^{p,q}=H_h^p(H_v^{q}D)$. Filtering by the second index gives the spectral sequence taking horizontal cohomology first and vertical cohomology next. Both filtrations are finite, so both converge to $H^{\ast }(\mathrm{Tot}D)$. This is the two spectral sequences of a bounded double complex construction.

The printed left/left tensor expression needs a handedness repair. Over an arbitrary ring, a tensor product of modules over $R$ pairs a right module with a left module. To retain the order of the printed formula, take $P$ to be a bounded cochain complex of projective right $R$-modules and $M$ a left $R$-module. Alternatively, keep $P$ left, take $M$ right, and write $M{\otimes }_{R}P$ and ${\mathrm{Tor}}_{-p}^R(M,H^{q}P)$. For a commutative ring no repair is needed. We prove the first, correctly typed formulation.

Finite projective dimension gives a finite projective resolution $Q_{\bullet }\to M$ by left modules. Form $D^{p,q}=P^q{\otimes }_R{Q}_{-p}$ for $-d\le p\le 0$. On $x\otimes y$ take $d_v=d_P\otimes 1$ and $d_h=(-1{)}^{q}1\otimes d_Q$. These anticommute, and the double cochain complex is bounded in both directions. Projective modules are flat modules, so taking vertical cohomology first givesThe first identification uses flatness of $Q_{-p}$; the second is the Tor functor computed by resolving its left-module argument $M$. Balancedness of the Tor functor allows either correctly sided projective resolution to compute it, by the double-resolution argument.

Taking horizontal cohomology first instead uses flatness of $P^q$. The resolution $Q_{\bullet }$ then leaves only $P^q{\otimes }_{R}M$ in column $p=0$. Its remaining differential is $d_P\otimes 1$, so the augmentation $\mathrm{Tot}D\to P{\otimes }_{R}M$ is a quasi-isomorphism. Apply the convergence theorem to this underlying abelian-group double complex. The two spectral sequences therefore prove the Künneth spectral sequence:These are abelian groups in general; an extra module structure requires appropriate bimodule hypotheses. The index $p$ is nonpositive, so $-p$ is the nonnegative homological index of the Tor functor.

Finally write the cycle modules as $Z^q=\ker d^q$ and the boundary modules as $B^q=\mathrm{im}{d}^{q-1}$. If every boundary module $B^q$ is projective, the short exact sequence $0\to Z^q\to P^q\to B^{q+1}\to 0$ splits. Thus $Z^q$ is projective. The short exact sequence $0\to B^q\to Z^q\to H^{q}P\to 0$ is now a length-one projective resolution, so ${\mathrm{Tor}}_i^R(H^{q}P,M)=0$ for $i\ge 2$. The $E_2$ page occupies only columns $p=-1,0$. Every $d_r$ for $r\ge 2$ goes $r$ columns to the right, hence has zero source or zero target. ThereforeThis two-column degeneration of a Künneth spectral sequence gives the edge short exact sequencesDegeneration itself does not supply a canonical splitting. Projective boundaries also do not force $H^{q}P$ to be projective: a complex with differential multiplication by $2$ over $\mathbb{Z}$ has projective boundary but a $\mathbb{Z}/2\mathbb{Z}$ cohomology group.

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.