Any hyperelliptic curve is never a complete intersection.

Question

Show that any hyperelliptic curve is never a complete intersection. As any curve of genus greater than 1 is either hyperelliptic or canonical, I think we can equivalently show that any curve of genus greater than 1 which is a complete intersection must be canonical. Any ideas?

Answer 1

Suppose that the curve $C$ has genus $g> 1$ and it is a complete intersection of hypersurfaces $X_1,\dots ,X_{n-1}$ in $\mathbb{P}^n$. We will calculate the canonical sheaf by induction. By adjunction formula, $$\omega_{X_1}=\omega_{\mathbb{P}^n}\otimes O_{X_1}(\deg X_1)=O_{\mathbb{P}^n}(-n-1)\otimes O_{X_1}(\deg X_1)=O_{X_1}(\deg X_1-n-1)$$ Also $$\omega_{X_1\cap X_2}=\omega_{X_1}\otimes O_{X_1\cap X_2}(\deg X_2)=O_{X_1\cap X_2}(\deg X_1+\deg X_2-n-1)$$ Proceednig like that we get $$\omega_C=O_C(\deg X_1+\dots+\deg X_{n-1}-n-1)$$ Since $\deg \omega_C=2g-2>0$ the expression $\deg X_1+\dots+\deg X_{n-1}-n-1$ must be positive so the canonical sheaf is very ample and $C$ cannot be hyperelliptic.

Answer 2

As suggested in the question, it suffices to show that, if $C$ is a complete intersection of genus greater than 1, then $K_C$ is very ample. Let $i_C: C\hookrightarrow\mathbb{P}^n$ be the complete intersection of hypersurfaces of degrees $d_1, \cdots, d_{n-1}$. First off,

Claim: For any $P, Q\in C$, $h^0(C, L(P+Q))=1$.

Proof: Equivalently, it suffices to show that the linear system $L(P+Q)$ consists of constant functions only. Suppose on the contrary that $f\in L(P+Q)$ is not a constant rational function. Write $\displaystyle f=\frac{i_C^*g}{i_C^*h}$, $g, h\in\Gamma\mathcal{O}_{\mathbb{P}^n}(m)$ for some $m\geq 1$. The possible zeros of $i_C^*h$ are $P$ or $Q$ or both of them. Let $S$ be a degree $m$ hypersurface defined by $h$. Then $S\cap C=\{P\}, \{Q\}$, or $\{P, Q\}$ and thus by Bezout theorem, $|S\cap C|=md_1\cdots d_{n-1}=1$ or $2$. If $|S\cap C|=1$ then $d_1=\cdots=d_{n-1}$ and $C\cong\mathbb{P}^1$, which is of genus 0. If $|S\cap C|=2$ either $m=2$ and $d_1=\cdots=d_{n-1}$, in which case $g(C)=0$, or one of $d_i=2$ and $d_j=1$ for $j\neq i$, in which case $C$ is isomorphic to a conic (i.e. $\mathbb{P}^1$), which again has genus 0. So we have contradiction.

By Riemann-Roch, \begin{align*} &h^0(C, L(P+Q))-h^0(C, K_C\otimes L(-P-Q))=3-g(C)\\ \Longrightarrow &h^0(C, K_C\otimes L(-P-Q))=g(C)-2=h^0(C, K_C)-2 \end{align*} The last line is exactly equivalent to $K_C$ being very ample.