Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 3

For a smooth projective curve $C$ of geometric genus $g$ and a divisor on an algebraic curve $D$, write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ and let $K_C$ be a canonical divisor. The curve form of the Riemann-Roch theorem isHere $\mathrm{deg}{K}_C=2g-2$ and $\ell (K_C)=g$. In line bundle form,The usual curve convention is a smooth geometrically integral projective curve; geometric embedding tests may be checked after extending the ground field to an algebraic closure.

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.

Let $\mathcal{L}={\mathcal{O}}_C(1)$. Its degree is five. Because the linear span of $C$ has dimension at least three, restrictions of linear forms supply at least four independent global sections, soIf $h^1(C,\mathcal{L})=0$, the Riemann-Roch theorem gives $h^0(\mathcal{L})=6-g$, and therefore $g\le 2$ immediately.

If instead $\mathcal{L}$ is special, the Clifford inequality for curves givesso its integer dimension is at most three, a contradiction. Here is a short proof of the required Clifford inequality for curves, so the bound need not be assumed. Put $r=h^0(\mathcal{L})>0$ and $s=h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$. At a smooth geometric point, choose bases of these two section spaces with strictly increasing orders of vanishing $a_1<\cdots <a_r$ and $b_1<\cdots <b_s$. Such bases exist because each successive order-of-vanishing quotient has dimension at most one. The products of the first basis with the first element of the second basis, followed by products of the last element of the first basis with the remaining elements of the second, have ordersThey are linearly independent global sections of ${\omega }_C$, proving the product dimension bound for sections on a curve. Thus $r+s-1\le h^0({\omega }_C)=g$. Combining this with Riemann-Roch theorem, $r-s=\mathrm{deg}\mathcal{L}+1-g$, gives $2r\le \mathrm{deg}\mathcal{L}+2$, which is exactly the displayed Clifford inequality for curves.

The special case is impossible, and henceThis is the genus bound for a nonplanar degree-five curve. The geometric argument can be carried out after extension to an algebraic closure, which preserves these section dimensions.