TITLE: Any hyperelliptic curve is never a complete intersection.
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.