Let $k$ be an algebraically closed field with char $k \neq 2$. Let $C$ be the affine plane curve in $\mathbb{A}^{2}$ defined by: $y^{2}=f(x)$, where $f(x)=c_{d}x^{d}+c_{d-1}x^{d-1}+...+c_{0} \in k[x]$ has no repeated roots, $c_{d} \neq 0$.
Let $C'$ be the unique nonsingular projective curve birational to $C$. The map $\phi: C \to \mathbb{A}^{1}$ by $(x,y) \mapsto x$ induces a morphism $\pi:C' \to \mathbb{P}^{1}$. Use $\pi$ to compute the genus of $C'$ in terms of $d$.
I am trying to apply Hurwitz formula, we have $d$ ramification points, each of index $2$. So the degree of ramification divisor is d. Now what is the degree of $\pi$? If $\phi$ is of degree $1$, can we say $\pi$ is also degree $1$?
If degree is indeed $1$, then I got $g=\frac{d}{2}$, which contradicts to the result from the Hilbert polynomial, which says if $C$ is a plane curve of degree $d$, $g=\frac{(d-1)(d-2)}{2}$.
Can someone explain this? Thanks a lot!