TITLE: Can any smooth hyperelliptic curve be embedded in a quadric surface?
QUESTION [8 upvotes]: Let's consider algebraic curves over a fixed algebraically closed field $K$.
It's well known, that every smooth elliptic curve (genus $g = 1$) can be embedded in a quadric surface in $\mathbb{P}^3$. This fact follows simply from the Riemann–Roch theorem.
More generally, for smooth hyperelliptic curves of higher genus ($g \ge 2$) it's known that such curves can be embedded in a quadric in weighted projective space $\mathbb{P}(1,1,g)$, see, for example, work of D. Eisenbud. (So, in the case $g=2$ we have embedding in $\mathbb{P}^4$).

But, can any smooth hyperelliptic curve $H$ be embedded in a quadric surface in $\mathbb{P}^3$?

It's natural question, because, by the definition we have mophfism $\phi:H \to \mathbb{P}^1$ of degree 2.
I think, this problem is connected with topics "Families of hyperelliptic curves and double covers of quadric surface" and Quotient Surface of A Hyperelliptic Involution, but I don't get it. (I'm interested in the case of any algebraicaly closed field and in the case of finite field).

REPLY [11 votes]: An explicit realization of degree $2$ and degree $g+1$ maps that separate points can be provided. Suppose the equation of a hyperelliptic curve is $$C:y^2=f(x)$$ with $\deg(f)=2g+2$. "Complete the square" by writing $$f(x)=r(x)^2+q(x)$$ with $\deg(r)=g+1$ and $\deg(q)\leq g$. Then, the maps $$A:(x,y) \mapsto x$$ $$B:(x,y)\mapsto y-r(x)$$ are degree $2$ and degree $g+1$ respectively. The map $B$ is degree $g+1$ because if we assume that $B(x,y)=c$, then we get the equation $c(c+2r(x))=q(x)$, which generically has $g+1$ solutions. Furthermore $(A,B)$ is clearly injective. So, the image $$(A,B):C\mapsto C'\subset\mathbb{P}^1\times\mathbb{P}^1$$ is a degree $(2,g+1)$ curve that $C$ normalizes and the logic of Jack's answer applies.