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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 $141$.

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.

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.

For a coherent sheaf $\mathcal{F}$ of support dimension $d$ on a projective scheme and a fixed Cartier divisor $D$, $h^0(X,\mathcal{F}(mD))=O(m^d)$. Choose a sufficiently high ample divisor $A$ whose section avoids the associated points of $\mathcal{F}$ and for which $D+A$ is ample. Multiplication by the $m$th power of that section injects $\mathcal{F}(mD)$ into $\mathcal{F}(m(D+A))$. The ample Hilbert polynomial, together with Serre vanishing, bounds the latter section space by $O(m^d)$.