Ramification points and Weierstrass points This is question 7.4.7 out of Liu's book:
Let X be a hyperelliptic curve of genus $g \geq 2$ endowed with a separable morphism $f: X \rightarrow \mathbb{P}^1_k$ of degree 2. We can write $K(X) = k(t)[y]$ with a relation $y^2+Q(t)y=P(t)$ .Let $x_0 \in X(k)$ and assume that $x_0$ is a Weierstrass point, i.e that $l(2x_0) \geq 2$. We want to show that a Weierstrass point is a ramification point. We know that fixed points of the hyperelliptic involution $\rho$ are ramification points.
The exercise is divided in steps, in a) assume that $x_0$ is not a ramifcation point, then $x'_0 = \rho(x_0)$ is distinct from $x_0$. Now you take $h \in L(2x_0) \setminus k$ , and show that $(h \pm \rho(h))_\infty = 2[x_0]+2[x'_0] = 2 f^\ast [f(x_0)]$. I have no problem with showing this. 
In b) you show that you can assume that h is of the form $h=a(t)+b(t)y$ . I can do this without problem. Here is my problem:
c) By considering the degree of $h - \rho(h)$, show that $g \leq 1$.
How can I calculate the degree of $h - \rho(h)$ and how can I from it, conclude that it has genus less than or equal to 1?
 A: I think the point $f(x_0)$ is $t=\infty$. 
If $f$ is unramified at $x_0$, then $t$ is a parameter at $x_0$ : 
$\mathrm{ord}_{x_0}t=1$. 
Let $g$ be the genus of $X$. Then $\deg P=2g+1$ or $2g+2$ and $\deg Q\le g+1$. 
We have $\sigma(y)=-y-Q$. So $h-\sigma(h)=b(t)(2y+Q(t))$. Now distinguish two cases. 


*

*If $k$ has characteristic $2$, then 
$$2=\deg b(t)+\deg Q(t)\ge \deg Q(t).$$ 
As $f$ is unramified above $t=\infty$, and the double cover $f$ around $\infty$ is given by the equation
$$(\frac{y}{t^{g+1}})^2+(\frac{Q(t)}{t^{g+1}})\frac{y}{t^{g+1}}=\frac{P(t)}{t^{2g+2}},$$ 
we must have $\frac{Q(t)}{t^{g+1}}|_{t=\infty}\ne 0$. This means $\deg Q(t)=g+1$. The above inequality then implies that $g\le 1$. 

*If $k$ has characteristic different from $2$. We can choose $Q=0$. Then
$$(\frac{y}{t^{g+1}})^2=\frac{P(t)}{t^{2g+2}}.$$
As $f$ is supposed to be unramified above $\infty$, the RHS must be non-zero at $t=\infty$. In other words, $\deg P(t)=2g+2$ and $y/t^{g+1}$ is regular and invertible at points above $\infty$. So $\mathrm{ord}_{x_0}(y)=g+1$. On the other hands, 
$$2=\mathrm{ord}_{x_0}(h-\sigma(h))=\deg b(t)+ \mathrm{ord}_{x_0}(y)\ge g+1.$$ 
Hence $g\le 1$. 
