Show every hyperelliptic curve admits a morphism of degree 2

Question

I am given an alternative definition for a hyperelliptic curve as follows. A hyperelliptic curve is a smooth projective curve $C$ over an algebraically closed field $\bar{F}$ with a divisor $D\in \operatorname{Div}(C)$ with $\operatorname{deg}(D)=2$ and $\operatorname{dim}\mathcal{L}(D)\geq 2$, where $\operatorname{dim}\mathcal{L}(D) = \{ f\in\bar{F}(C)^* : div(f)+D\geq 0\} \cup \{0\}$.

I have to show every hyperelliptic curve admits a morphism $C\to \mathbb{P}^1$ of degree 2.

My idea: since the dimension of the linear system is at least 2, I can pick linearly independent functions $f,g$ and consider the map $P\mapsto [1:f(P)/g(P)]$. However, I am unsure how to show this map is of degree 2. I'm sure I have to use the conditions on $f,g$ somewhere, but I'm not sure how to approach this, especially showing this map is of degree 2.

Answer 1

You have the good idea. However, be careful with your notations, $[1:f(P)/g(P)] \in \mathbb{P}^1$ has sens only when $f(P)/g(P) \in F$. In fact, you can see $\mathbb{P}^1$ as $F \cup \{\infty\}$ just like for the Riemann sphere. Then, $F(C)$ is simply the set of morphisms $C \rightarrow \mathbb{P}^1$ and it has a field structure so you can directly consider $h : P \mapsto \frac{f(P)}{g(P)} \in F(C)$.

$(f,g)$ linearly independant implies that $h$ is non constant and the degree of $h$ equals the cardinality of its fibers counted with multiplicities. In particular, the number of zeros of $h$ plus its number of poles (still with multiplicities) is twice its degree.

Notice that for all $P$, \begin{align*} h(P) = 0 & \Rightarrow \mathrm{div}(h)_P > 0 \textrm{ and } h \textrm{ has a zero of order } \mathrm{div}(h)_P \textrm{ at } P.\\ h(P) = \infty & \Rightarrow \mathrm{div}(h)_P < 0 \textrm{ and } h \textrm{ has a pole of order } -\mathrm{div}(h)_P \textrm{ at } P.\\ h(P) \in F^* & \Rightarrow \mathrm{div}(h)_P = 0. \end{align*} We deduce form it that, $$ 2\deg(h) = \textrm{ number of zeroes + number of poles } = \sum_{P \in C} |\mathrm{div}(h)_P|. $$ And for all $P$, \begin{align*} |\mathrm{div}(h)_P| & = |\mathrm{div}(f)_P - \mathrm{div}(g)_P|\\ & \leqslant |\mathrm{div}(f)_P + D_P| + |-D_P - \mathrm{div}(g)_P|\\ & = \mathrm{div}(f)_P + \mathrm{div}(g)_P + 2D_P. \end{align*} By summing over $P$, we obtain that, $$ 2\deg(h) \leqslant 2\deg(D) = 4. $$ Therefore, $h$ has degree at most two. If $h$ has degree one, replace it by $h^2$ (and in this particular case, $C = \mathbb{P}^1$ because $h$ is an isomorphism).

Answer 2

little informal remark how I think about this: take the map $\phi: C \rightarrow \mathbf{P}^1$ induced by taking two global sections $f,g \in H^0(\mathcal{L}(D))$. Then $\phi^*\mathcal{O}(1)\cong \mathcal{L}$. In terms of divisors, this is $\phi^*H=D$ where $H$ denotes a "hyperplane class", i.e. is the class of a single point $p\in \mathbf{P}^1$. But $D$ is of degree 2, i.e. linear equivalent to two points $q, q'$ (or $1$ point $2q$ with multiplicity $2$). As pullback is like inverse image, you get exactly the two points $q,q'$ are mapped to $p$ via $\phi$.

$\mathbf{Remark}$: This is generic situation, could be that you pullback only to a thickened point $2q$ but that still gives you a degree $2$ map although the reduced fibre is a singleton.