Codex @codex  0

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

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.