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The degree of a projective algebraic variety of dimension $r$ is the length, with intersection multiplicities, of its intersection with a sufficiently general linear subspace of complementary codimension. Equivalently, its Hilbert polynomial has leading coefficient $\mathrm{deg}X/r!$. For a projective curve, a general hyperplane section measures this degree.

For a projective complete intersection curve cut out by a homogeneous regular sequence of positive degrees $a,b$ in ${\mathbb{P}}_k^3$, the Koszul resolution gives Hilbert polynomial $abm+ab(4-a-b)/2$. Hence its arithmetic genus is $1+ab(a+b-4)/2$. The same resolution shows $H^0(C,{\mathcal{O}}_C)=k$, and the dimension of a scheme is one, so this genus equals ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Smoothness and irreducibility of the curve are not required. For degrees five and seven, the value is $141$.

For a projective scheme $X$ over a field $k$, the Hilbert functor assigns to a $k$-scheme $T$ the set of closed subschemes $W\subseteq X{\times }_{k}T$ which are of finite presentation, proper and flat over $T$. Pullback of families makes it a contravariant functor. After choosing a very ample line bundle, requiring a fixed fibre Hilbert polynomial $P$ gives the subfunctor represented by ${\mathrm{Hilb}}^P(X)$. The unrestricted functor is represented by the disjoint union of these Hilbert schemes.

For a projective scheme $X$ and a fixed Hilbert polynomial $P$, the Hilbert scheme parametrizes flat families of closed subschemes of $X$ with that polynomial. This turns a family of embedded algebraic curves into an algebraic parameter space.

A Cartier divisor $D$ on an integral projective variety is big exactly when, for every ample divisor $A$, some positive integer $j$ satisfies $jD\sim A+E$ with $E$ effective. To prove the forward direction, subtract an effective high multiple of $A$ using the section subtraction lemma for big divisors, then add an effective representative of the remaining multiple. For the reverse direction, multiply sections of $qA$ by the section of $qE$ and use the ample Hilbert polynomial.

Let $S=k[x_0,\dots ,x_n]$ and let $B=S/I(X)$ be the homogeneous coordinate ring of a nonempty projective variety. Its Hilbert function is $h_X(t)={\dim }_k{B}_t$. For all sufficiently large integers $t$, this function agrees with a polynomial $P_X(t)$, the Hilbert polynomial.

The dimension $r$ is the largest length of a strict chain of nonempty irreducible closed subvarieties of $X$. Equivalently $r=\dim B-1=\mathrm{deg}{P}_X$. The degree of the embedded variety is the integer specified by the leading termGeometrically it is the number of intersection points with a general linear subspace of codimension $r$, counted with intersection multiplicities. The polynomial's degree is the dimension $r$; the variety's degree is the leading coefficient multiplied by $r!$. Keeping those two meanings separate is important in the next computation.

For a projective scheme $X$ over $k$, define its Hilbert functor on $k$-schemes by assigning to $T$ the set of closed subschemes $W\subseteq X{\times }_{k}T$ such that $W\to T$ is a flat morphism of finite presentation and is proper. For projective $X$, the properness is automatic for these closed families. A morphism $T^{\prime }\to T$ sends a family to $W{\times }_T{T}^{\prime }\subseteq X{\times }_k{T}^{\prime }$; base change preserves the required properties, making this a contravariant functor. The embedding in $X\times T$ is part of the data, so these are embedded families rather than families modulo automorphisms of $X$.

Choose a very ample line bundle on $X$. In a flat projective family of finite presentation the fibre Hilbert polynomial is locally constant. Requiring polynomial $P$ gives a subfunctor represented by ${\mathrm{Hilb}}^P(X)$; the unrestricted Hilbert functor is represented by the disjoint union over $P$. For the specified $Z$, use its polynomial $P_Z$ and let $h=[Z]$ be the resulting $k$-point of the Hilbert scheme $H$.

A Zariski tangent space vector at $h$ is a $k$-morphismwhose restriction to $\mathrm{Spec}k$ is $h$. Here $D$ is the ring of dual numbers. Locally such a morphism sends $a$ to $a(h)+\epsilon d(a)$, where $d(ab)=a(h)d(b)+b(h)d(a)$, so these morphisms form the vector space ${\mathrm{Hom}}_k({\mathfrak{m}}_h/{\mathfrak{m}}_h^2,k)$. By the representing property of the Hilbert scheme, the same tangent vector is exactly a first-order embedded deformationwith the last equality an equality of embedded closed subschemes.

Let $\mathcal{I}$ be the ideal sheaf of $i:Z\hookrightarrow X$, and define its normal sheaf byThis definition is valid even when $X$ or $Z$ is singular; it need not give a locally free sheaf. We now derive the tangent-space identification by constructing inverse maps, rather than merely invoking the normal sheaf's name.

Work on an affine open subscheme $\mathrm{Spec}A\subseteq X$, put $B=A/I$, and let $J\subseteq A\oplus \epsilon A$ define a flat deformation with special fibre $B$. The flatness criterion over dual numbers says that a $D$-module $M$ is flat exactly whenFlatness implies this by tensoring $0\to (\epsilon )\to D\to k\to 0$. Conversely lift a $k$-basis of $M/\epsilon M$ to $M$. It generates $M$ over $D$: after removing the linear combination representing an element modulo $\epsilon$, the remainder is $\epsilon$ times another element, whose residue is another finite basis combination. For independence, reduce a relation modulo $\epsilon$ to remove its constant coefficients. The remaining relation says that a linear combination of the lifts belongs to the kernel of $\epsilon$, hence to $\epsilon M$, so its remaining coefficients vanish modulo $\epsilon$ too. The lifts are a free $D$-basis, establishing flatness.

Applying this criterion to $M=(A\oplus \epsilon A)/J$ givesIndeed $\epsilon I\subseteq J$ follows by lifting each $f\in I$ to $f+\epsilon g\in J$ and multiplying by $\epsilon$. Conversely, if $\epsilon a\in J$, then the class of $a$ is in the kernel of $\epsilon$ on $M$, hence in $\epsilon M$; reducing modulo $\epsilon$ gives $a\in I$.

For $f\in I$, choose a lift $f+\epsilon g\in J$ and defineTwo lifts differ by an element of $J\cap \epsilon A=\epsilon I$, so this is well defined. Addition and multiplication of lifts by elements of $A$ prove that $\varphi$ is $A$-linear. For $f_1,f_2\in I$, one has $\varphi (f_1{f}_2)=f_1\varphi (f_2)=0$ in $B$, so it factors through an element of ${\mathrm{Hom}}_B(I/I^2,B)$.

Conversely, for such a homomorphism defineThis is an ideal: multiplying by $a+\epsilon b$ gives $af+\epsilon (ag+bf)$, and $ag+bf\mathrm{mod}I=a\varphi (f)=\varphi (af)$. Its reduction is $I$. To prove that its quotient is flat, suppose $\epsilon [a+\epsilon b]=0$. The condition $\epsilon a\in J_{\varphi }$ says $a\in I$. Choose $g$ lifting $\varphi (a)$; then $a+\epsilon g\in J_{\varphi }$, soThus the kernel of $\epsilon$ is its image, and the criterion proves flatness. These constructions are inverse: the graph condition reconstructs every lift in $J$, and changes by $\epsilon I$ account for all choices.

Localization respects both constructions. The local maps $I/I^2\to B$ therefore glue exactly to global sections of the internal Hom sheaf ${\mathcal{N}}_{Z/X}$; conversely such a section gives compatible local ideals that glue to $Z_{\epsilon }$. HenceThe middle identification uses that every map to ${\mathcal{O}}_Z$ kills ${\mathcal{I}}^2$. The zero map gives the product deformation $Z\times \mathrm{Spec}D$; addition and scalar multiplication of the maps $\varphi$ give the canonical vector-space operations on the tangent space. This proves the Zariski tangent space of a Hilbert scheme formula without a smoothness or regular-embedding hypothesis.

Use the graded ring homomorphismwhere source degree $m$ maps to target degree $3m$. The homogeneous ideal of the twisted cubic is $\ker \phi$: a homogeneous polynomial vanishes on the parametrized curve precisely when its substitution vanishes identically. Here $k$ is infinite, being an algebraically closed field, so vanishing at all parameter points implies the polynomial identity.

Let $J$ be generated byTheir substitutions vanish, hence $J\subseteq \ker \phi$. To prove the reverse inclusion as an equality of ideals, we need more than a set-theoretic description of the zero locus.

Modulo $J$, replace $x_1^2$ by $x_0{x}_2$, $x_2^2$ by $x_1{x}_3$, and $x_1{x}_2$ by $x_0{x}_3$. Each replacement strictly decreases the sum of the exponents of $x_1$ and $x_2$, so the reduction terminates. It shows that the degree-$m$ quotient is spanned by the standard monomials of the twisted cubicFor $m\ge 1$ there are $(m+1)+m+m=3m+1$ of these. For $m=0$ the sole monomial is $1$. Their substituted images have $t$-exponents respectively $3b$, $3b+1$ and $3b+2$, so they are exactly the distinct monomialsThese are linearly independent. Consequently any degree-$m$ polynomial in $\ker \phi$, after reduction modulo $J$, has all its reduced coefficients zero. It belongs to $J$. Applying this to each homogeneous component givesThis also proves that the quotient embeds in the domain $k[s,t]$, so the displayed ideal is prime; no radicalization or saturation correction is concealed in the argument. Its image is the third Veronese subring ${⨁}_{m\ge 0}k[s,t{]}_{3m}$, and the calculation gives the Hilbert polynomial $3m+1$, consistent with degree three and genus zero.

Let $X$ be a separated Noetherian scheme over a field $k$ and $\mathcal{F}$ a coherent sheaf. Coherent cohomology studies the groups $H^q(X,\mathcal{F})$: global sections in degree zero and the obstructions to gluing local sections in higher degrees. Coherence means finite generation locally, and on a Noetherian scheme it is preserved by kernels, cokernels and extensions. This makes coherent sheaves suitable for exact-sequence calculations.

For an ordered open cover $\mathcal{U}=(U_i)$, the Čech cochain groups arewith differentialEach double omission occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^q=\ker \delta /\mathrm{im}\delta$, and taking the direct limit over refinements defines cover-independent Čech cohomology. In degree zero the sheaf gluing axiom identifies it with $\Gamma (X,\mathcal{F})$.

The comparison with derived-functor sheaf cohomology is particularly effective algebraically. Positive-degree cohomology of a quasi-coherent sheaf on an affine scheme vanishes. A separated scheme has affine intersections of affine opens, so a finite affine cover is acyclic for $\mathcal{F}$. Its Čech cochain complex consequently computes $H^q(X,\mathcal{F})$ directly. For a short exact sequence of sheaves, sections over each affine intersection are exact on quasi-coherent sheaves. The resulting short exact sequence of Čech complexes gives the long exact sequence in sheaf cohomology. In particular, the connecting map measures the failure of a global section of the quotient to lift globally.

Several central projective results make the theory finite and computable. On a projective variety over $k$, every coherent cohomology group is finite-dimensional. The Grothendieck vanishing theorem gives $H^q(X,\mathcal{F})=0$ for $q>\dim X$. If ${\mathcal{O}}_X(1)$ is ample, Serre vanishing gives $H^{q>0}(X,\mathcal{F}(m))=0$ for sufficiently large $m$, and these large twists are globally generated sheaves. On projective space a coherent sheaf has a finite twisting resolution of a coherent sheaf on projective space: finite sums of $\mathcal{O}(a)$ resolve it. The explicit line-bundle cohomology below and the long exact sequences prove finiteness and positive-twist vanishing. For a closed embedding $i:X\hookrightarrow {\mathbb{P}}^n$, $i_{\ast }\mathcal{F}$ is coherent and $H^q(X,\mathcal{F})=H^q({\mathbb{P}}^n,i_{\ast }\mathcal{F})$, reducing these results to projective space. A sufficiently positive twist of a finite presentation by sums of $\mathcal{O}(-a)$ also proves global generation.

The Euler characteristic of a coherent sheafis therefore defined, and the long exact sequence makes it additive. With a fixed projective embedding, $\chi (X,\mathcal{F}(m))$ is its Hilbert polynomial, and for sufficiently large $m$ it equals $h^0(X,\mathcal{F}(m))$. On a smooth projective variety of dimension $r$, Serre duality gives, for a vector bundle $E$,For an arbitrary coherent sheaf the correct counterpart is $H^q(X,\mathcal{F}{)}^{\ast }\cong {\mathrm{Ext}}_X^{r-q}(\mathcal{F},{\omega }_X)$; replacing this Ext group indiscriminately by the cohomology of the ordinary sheaf dual would lose local extension information.

We now calculate the basic examples. For $n\ge 1$, use the standard cover $U_i=D_{+}(x_i)$ of ${\mathbb{P}}^n$. Sections of $\mathcal{O}(m)$ on an intersection are degree-$m$ Laurent monomials in which negative exponents are allowed only for coordinates inverted on that intersection. Decompose the Čech cochain complex by exponent vectors $(a_0,\dots ,a_n)$ with sum $m$. If no exponent is negative, the corresponding simplex complex contributes only in degree zero. If all exponents are negative, the monomial occurs only on the full intersection and contributes only in degree $n$. When the negative-index set is nonempty and proper, inserting any index outside it gives a contracting homotopy, so the monomial complex is acyclic. Hence the cohomology of twisting sheaves on projective space isIn the top group a basis consists of the Laurent monomials with every exponent at most $-1$. Writing $a_i=-1-b_i$ reduces the count to nonnegative solutions of $\sum b_i=-m-n-1$, giving the displayed binomial coefficient. In particular, all intermediate cohomology vanishes, and ${\omega }_{{\mathbb{P}}^n}=\mathcal{O}(-n-1)$ agrees with Serre duality. For ${\mathbb{P}}^0$ separately, every twist is trivial and $H^0=k$.

On the projective line this givesThus positive and negative degrees produce quite different groups, while their alternating difference is a single polynomial. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and its hyperplane bundle corresponds to ${\mathcal{O}}_{{\mathbb{P}}^1}(3)$. Consequently $h^0(X,{\mathcal{O}}_X(m))=3m+1$ for $m\ge 0$, agreeing with Q2, and $h^1(X,{\mathcal{O}}_X(m))=\max (-3m-1,0)$.

For a smooth plane curve $C$ of degree $d\ge 1$, the ideal-sheaf sequence of a projective hypersurface isThe long exact sequence in sheaf cohomology and the preceding formula give $H^0(C,{\mathcal{O}}_C)=k$ and $H^1(C,{\mathcal{O}}_C)\cong H^2({\mathbb{P}}^2,\mathcal{O}(-d))$. ThereforeFor $d=1,2$ this is zero; for a plane cubic it is one. Twisting the same exact sequence gives $\chi (C,{\mathcal{O}}_C(m))=dm+1-g$. These examples illustrate how Čech cohomology, vanishing, Serre duality and exact sequences convert local equations into global numerical invariants.

Work over $\mathbb{C}$ in characteristic zero. The second Weyl algebra has generators $x_1,x_2,{\partial }_1,{\partial }_2$ and relations $[{\partial }_i,x_j]={\delta }_{ij}$, $[x_i,x_j]=[{\partial }_i,{\partial }_j]=0$. Its Bernstein filtration gives every generator degree one. The ordered monomials $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}{\partial }_2^{b_2}$ form a basis, soFor a nonzero finitely generated module, choose a finite-dimensional generating subspace $M_0$ and put $M_r=F_r{A}_2{M}_0$. Its graded module is finite over this polynomial ring. The Hilbert polynomial theorem gives an eventual polynomial $\dim M_r$, whose degree $d(M)$ is the Bernstein growth dimension of a Weyl algebra module, equal to its Gelfand–Kirillov dimension of a module. Two choices of finite generators bound each other after fixed filtration shifts, proving independence of the choice. The characteristic variety of a Weyl algebra module is the support of this graded module; its algebraic dimension is $d(M)$. In particular $d(M)$ is an integer at most four.

Here is a growth proof of the lower bound, rather than just quoting the Bernstein inequality for Weyl algebra modules. A nonzero $u\in F_r{A}_2$ can be turned into a nonzero scalar by at most $r$ successive commutators with the four generators: choose a monomial of maximal total degree, differentiate it to a constant using $[{\partial }_i,\cdot ]$ and $[x_i,\cdot ]$, and all different monomials are killed by that derivative multi-index. This uses the ordered-monomial expressions and characteristic zero. Expanding the iterated commutator expresses that scalar as ${\sum }_j{v}_{j}u{w}_j$ with $v_j,w_j\in F_r{A}_2$.

If $u$ acted as zero on $M_r$, every $u{w}_j{M}_0$ would be zero, so the scalar would kill the nonzero $M_0$, a contradiction. Therefore the action mapis injective. This is the faithful finite-step action of a Weyl algebra. Comparing growth ingives $4\le 2d(M)$. Thus the only possible dimensions for nonzero modules are $\boxed{d(M)\in \{2,3,4\}}$.

All three values occur in cyclic modules. On $\mathbb{C}[x_1,x_2]$, let $x_i$ multiply and ${\partial }_i$ differentiate. The module is generated by $1$, its filtered pieces are precisely polynomials of degree at most $r$, and $\dim M_r=\left(\genfrac{}{}{0ex}{}{r+2}{2}\right)$. Hence $\boxed{d(\mathbb{C}[x_1,x_2])=2}$; equivalently it is $A_2/(A_2{\partial }_1+A_2{\partial }_2)$.

For $M=A_2/A_2{\partial }_2$, the ordered basis shows that the left ideal $A_2{\partial }_2$ is spanned by monomials with positive ${\partial }_2$ exponent. The remaining basis is $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}$. Its degree-$\le r$ piece has dimension $\left(\genfrac{}{}{0ex}{}{r+3}{3}\right)$, so $\boxed{d(A_2/A_2{\partial }_2)=3}$. Finally the regular module has the four-variable ordered basis above, giving $\boxed{d(A_2)=4}$.

The dimension-two modules are the holonomic Weyl algebra modules. Exact sequences satisfy $d(M)=\max \{d(M^{\prime }),d(M^{\prime \prime })\}$, by using compatible good filtrations and the resulting Hilbert polynomials. Therefore nonzero subquotients of a holonomic module remain holonomic. Their positive integer multiplicities, namely $2!$ times the leading Hilbert-polynomial coefficient, add in exact sequences. Every nonzero successive factor consumes at least one unit of multiplicity, bounding the length of a submodule chain; hence holonomic modules have finite composition length. The polynomial module above is even simple: differentiating a nonzero polynomial to a constant and then multiplying by arbitrary polynomials generates the whole module. In contrast the regular and dimension-three examples exhibit the larger growth allowed for nonholonomic modules. These examples and the growth bound explain both the range and its representation-theoretic significance.

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.

Eliminate $t_3$ using the linear equation. The homogeneous coordinate ring becomesThe cubic is a nonzero element of a polynomial integral domain, hence a non-zero-divisor. It gives the exact sequence of graded modulesThus $\dim B_m=\left(\genfrac{}{}{0ex}{}{m+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$ for $m\ge 3$. Equivalently, use the Hilbert series $(1-z^3)/(1-z{)}^3$, or the sheafified exact sequence on ${\mathbb{P}}^2$. The Hilbert polynomial of the plane cubic isIts degree is three and its arithmetic genus is one, although the curve is singular: its affine equation near $[1:0:0]$ is the cusp $t_2^2+t_1^3=0$. The Hilbert polynomial records the arithmetic genus, rather than the genus of the normalization.

Fix any ample divisor $A$. Choose $q\gg 0$ so that both $qA$ and $(q-1)A$ have nonzero global sections, and choose $F\in |qA|$. Condition (1) and the section subtraction lemma for big divisors yield an integer $j>0$ with $jD-qA\sim E_0\ge 0$. Adding an effective member of $|(q-1)A|$ gives $jD\sim A+E$. Thus (1) implies the stronger condition (2').

If $jD\sim A+E$ with $A$ ample and $E$ effective, multiplication by the section of $qE$ injects $H^0(X,qA)$ into $H^0(X,qjD)$. The positive leading coefficient of the ample Hilbert polynomial supplies condition (1) along the infinite sequence $m=qj$. Condition (2) implies (3). Conversely, if $jD\equiv A+E$, put $P=jD-E$. The Cartier divisor $P$ is numerically equivalent to $A$, hence ample by Kleiman's criterion; the actual equality $jD=P+E$ gives (2).

We have proved (1)$\Rightarrow$(2')$\Rightarrow$(2)$\Rightarrow$(1), and (3)$\Rightarrow$(2)$\Rightarrow$(3). Also (2') implies (3'), while (3') implies (3). Hence all five conditions are equivalent. They characterize a big divisor; the ample-plus-effective formulation is Kodaira's lemma. The inconsistent use of $n$ and $j$ in the printed multiplier is resolved by using $j$ throughout.

Choose a very ample divisor $A$ such that $A-D$ has a nonzero global section; this is possible by taking a sufficiently high ample twist. Since $X$ is integral, multiplication by its $m$th power injects $H^0(X,mD)$ into $H^0(X,mA)$. By Serre vanishing and the Hilbert polynomial, $h^0(X,mA)=O(m^n)$. ConsequentlyThis is the polynomial bound for sections of a fixed divisor. The same proof works for a coherent sheaf of support dimension $d$, by choosing the multiplying section to avoid its associated points; the bound is then $O(m^d)$.

Assume first that $f\ne 0$ and letbe the least total degree of a nonzero homogeneous part of $f$. The associated graded ring iswhere $f_d$ is the initial homogeneous form. Multiplication by the nonzero polynomial $f_d$ is injective in $k[x,y]$, so the degree-$j$ component has dimensionSumming the components of degrees below $n$ givesThus the Hilbert polynomial isIts leading coefficient is the order of vanishing, or multiplicity, of the plane curve $f=0$ at the origin. If $f=0$, no relation is imposed and $\chi =n(n+1)/2$ for every $n$.

A Hilbert polynomial of $A$ is a polynomial $h_A(X)\in \mathbb{Q}[X]$ such thatfor every sufficiently large integer $n$. Eventual equality makes this polynomial unique.

For every integer $t$,For $t\ge 0$ this is the dimension of the homogeneous polynomials of degree $t$, since the higher cohomology vanishes; polynomiality then identifies the Hilbert polynomial.

If $P_X(t)=\chi (X,{\mathcal{O}}_X(t))$ is the Hilbert polynomial of a pure $k$-dimensional closed subscheme, thenEquivalently, if $H=c_1(\mathcal{O}(1))$, the degree of a projective scheme isA generic complementary linear subspace meets $X$ in a nonempty zero-dimensional scheme whose length is this number. It is therefore a positive integer.

Letbe the homogeneous coordinate ring. Its Hilbert function agrees for all sufficiently large $m$ with a polynomial $P_V(m)$. Since $V$ is a curve,The degree of a projective curve is the positive integer $d$. This is well-defined because the graded ring $A_V$, and hence its eventual Hilbert polynomial, depends only on the embedded projective variety, while a polynomial agreeing with the Hilbert function for all large integers is unique.

This definition also has the geometric interpretation expected of degree. For a sufficiently general hyperplane $H=Z(\ell )$ that does not contain $V$, multiplication by $\ell$ gives an exact sequenceConsequently, for large $m$,The left side is the length of the zero-dimensional scheme $V\cap H$. Thus a generic hyperplane meets $V$ in $d$ points counted with intersection multiplicity, independently of the chosen generic hyperplane.

For the stated linear embedding, the homogeneous coordinate ring of $\phi (V)$ isThe graded rings have the same Hilbert polynomial, so degree under a linear projective embedding gives

Degree is nevertheless not an invariant of an abstract curve. In ${\mathbb{P}}^2$, a line has degree one and the smooth conichas degree two, but the mapidentifies the conic with ${\mathbb{P}}^1$, as is any projective line.

For the specified set $S$, the equations are the $2\times 2$ minors ofOn the affine chart $x_0=1$, they forceso this chart is an affine line. When $x_0=0$, the equations force $x_1=x_2=0$, leaving the single point $[0:0:0:1]$. Hence $V$ is the twisted cubic, parametrized byIt is therefore an irreducible projective curve. A generic hyperplane pulls back to a binary cubicwhich has three zeros on ${\mathbb{P}}^1$ counted with multiplicity. Thus

No nonzero linear form vanishes on $V$, since substituting its parametrization would make the four coefficients of the displayed binary cubic vanish. Therefore every hypersurface containing $V$ has degree at least two. Ifthen $r=1$ is impossible because a nonzero hypersurface in ${\mathbb{P}}^3$ has dimension two. Hence at least two nonconstant $F_i$ are needed, andThis is the degree obstruction to the twisted cubic being a complete intersection.

Finally compare the twisted cubic with a smooth plane cubic, such asviewed in ${\mathbb{P}}^3$ by the given linear embedding. Both curves are irreducible and have degree three. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and has genus zero, whereas the supplied plane-curve genus formula givesSince genus is an isomorphism invariant, these equal-degree curves are not isomorphic.

A plane cubic is a projective plane curve defined by a nonzero homogeneous polynomial of degree three. Its arithmetic genus is one by the genus-degree formula, and its Hilbert polynomial is $3m$. A singular cubic can have normalization of smaller genus; the Hilbert polynomial measures the arithmetic genus.