Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 24F (Algebraic Geometry)

Let $H=V(L)$ for a nonzero homogeneous linear polynomial $L$. Because $H$ does not contain $X$, the restriction $L{|}_X$ is a nonzero section of ${\mathcal{O}}_X(1)$. Its zero divisor is the hyperplane-section divisor of a projective plane curvewhere $I_p(X,H)$ is the local intersection multiplicity; equivalently, it is the order of vanishing of $L{|}_X$ at $p$.

The curves $X$ and $H$ have no common component and have degrees $d$ and $1$. The Bézout theorem therefore gives

The Adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. Since a hyperplane section has degree $d$, the canonical degree of a smooth plane curve is

For any smooth projective curve of genus $g$, a canonical divisor has degree $2g-2$. If this plane curve had genus two, part (ii) would implyso $d^2-3d-2=0$. The quadratic discriminant is $17$, which is not a perfect square, so the equation has no integral solution. Thus a smooth projective plane curve cannot have genus two, as recorded by no smooth plane curve has genus two.

For a divisor on an algebraic curve $D$, define its Riemann-Roch space byThen

The inclusion $L((n-1)p)\subseteq L(np)$ allows one additional coefficient: the coefficient of $t^{-n}$ in the local Laurent expansion at $p$, where $t$ is a uniformizer. The resulting linear mapis injective, so the dimension can increase by at most one.

It increases precisely when some rational function has a pole of exact order $n$ at $p$ and no other poles. By the definition of the Weierstrass gap set $G(p)$, this occurs exactly when $n\notin G(p)$. Hence

For $N>2g-2$, the divisor $K_X-Np$ has negative degree, so $\ell (K_X-Np)=0$. The Riemann-Roch theorem givesStarting from $\ell (0)=1$ and using part (ii), each integer $1\le n\le N$ contributes one to $\ell (Np)$ exactly when it is not a gap. ThereforeComparison with the preceding formula gives $|G(p)\cap \{1,\dots ,N\}|=g$. Moreover every $n>N$ is a nongap because the Riemann--Roch formula makes every subsequent dimension jump equal to one. Thus the Weierstrass gap theorem says

First, $1\in G(p)$. Otherwise there would be a nonconstant rational function with one simple pole and no others, defining a degree-one finite morphism $X\to {\mathbb{P}}^1$. Such a map is an isomorphism, which would make $X$ have genus zero.

Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an additive semigroup under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the Genus-two Weierstrass gap sequence