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For a finite affine cover of a semi-separated scheme, every finite intersection is affine. The augmented sheaf Čech cochain complex is exact: near each point, an open of the cover contains the neighbourhood under consideration, and inserting its index contracts the complex. Its terms are acyclic for a quasi-coherent sheaf, by affine cohomology vanishing and the affine-intersection property. Alternatively, apply the section-level Čech cochain complex to a flasque resolution; the two computations of the resulting double complex identify Čech cohomology with sheaf cohomology. Stalks of direct images from opens outside those opens need not vanish, so exactness must not be justified by treating these direct images as extensions by zero.

The intersection hypothesis says that $X$ is a semi-separated scheme. For an affine open subscheme $V\subseteq X$, the intersection $V\cap U$ is affine. Restricting the given short exact sequence of sheaves to it and applying the affine module-sheaf equivalence gives an exact sequence of global sections:These are exactly the sections on $V$ of the three direct image sheaves. Affine opens form a basis, and every section of the last sheaf on such a basis open lifts to the middle sheaf. This proves surjectivity as a sheaf morphism; left exactness of direct image supplies the other positions. Hence direct image preserves this short exact sequence. The crucial ingredient is exactness of sections of quasi-coherent sheaves on an affine intersection; arbitrary open immersions need not have this property.

For the cohomology comparison, every nonempty finite intersectionis affine. This follows by mathematical induction, intersecting the affine intersection already obtained with the next affine open. The restriction of $\mathcal{F}$ to it is quasi-coherent, so vanishing of quasi-coherent cohomology on an affine scheme givesWe now prove why this local vanishing gives the acyclic cover theorem, rather than identifying the two sorts of cohomology without a comparison.

Take a flasque resolution $0\to \mathcal{F}\to {\mathcal{I}}^0\to {\mathcal{I}}^1\to \cdots$. Here a flabby sheaf has surjective restriction maps; its restrictions to open subsets are still flabby and have zero higher sheaf cohomology. Form the double complexThe horizontal differential is the alternating restriction map $d_{\mathrm{C}}$; the vertical one $d_{\mathrm{I}}$ is induced by the resolution. They commute, so the total differential in bidegree $(p,q)$ is $d_{\mathrm{C}}+(-1{)}^p{d}_{\mathrm{I}}$ and has square zero. The Čech resolution on a semi-separated scheme uses precisely these intersections.

A flabby sheaf has zero positive Čech cohomology for a finite open cover, and its degree-zero Čech cohomology is its global sections. One way to establish this auxiliary fact is to use the exact augmented two-open complexExactness at the first two terms is the sheaf gluing axiom; the last map is onto because a section on the intersection extends to $V$. For the induction step, write $V$ for the union of all but the last open and $W$ for the last open. Separate Čech cochains according to whether their index list contains the last index. The resulting two-block complex compares the smaller cover of $V$ with its restricted cover of $V\cap W$, together with $\Gamma (W,\mathcal{I})$ in degree zero. By induction those two smaller cover complexes have cohomology only in degree zero, where they give $\Gamma (V,\mathcal{I})$ and $\Gamma (V\cap W,\mathcal{I})$. The displayed two-open exact sequence then gives zero positive-degree cohomology for the full cover. This proves the auxiliary fact by induction on the number of opens. Thus horizontal cohomology of $C^{\bullet ,q}$ consists only of $\Gamma (X,{\mathcal{I}}^q)$ in degree zero. Computing the cohomology of the total complex first horizontally therefore gives $H^n(X,\mathcal{F})$, by the resolution principle for sheaf cohomology.

On the other hand, vertical cohomology issince the restricted flasque resolutions compute cohomology on each intersection. The already established affine vanishing makes all rows with $q>0$ zero. The surviving row $q=0$ is exactly the Čech cochain complex ${\check{C}}^{\bullet }(\mathcal{U},\mathcal{F})$. Computing total cohomology first vertically therefore gives ${\check{H}}^n(\mathcal{U},\mathcal{F})$. These two computations are justified by the two filtrations of the first-quadrant double complex: in every total degree only finitely many terms occur, and the cover also bounds the horizontal degree. Their edge maps give the natural identificationFor negative degrees both groups are zero by convention. In particular, degree zero is the usual identification by the sheaf gluing axiom, not merely a comparison of dimensions.

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 Tor functor is the homology of a tensor product of modules with a free resolution: ${\mathrm{Tor}}_i^R(M,N)=H_i(F_{\bullet }{\otimes }_{R}N)$ for a free resolution $F_{\bullet }$ of $M$. It measures the failure of tensoring to preserve exact sequences. For a commutative ring, one may resolve either argument; the double complex of the two resolutions proves that both computations agree.