This is exercise 2.2 in Hartshorne's AG:
What I have done:
a):Apply Hurwitz theorem and observe that ramification index is no greater than degree. b):There is $ k[x]\rightarrow k[x,z]/(z^2-\prod (x-\alpha_i ))$.So explicit calculation can be done. c):Follow the hints.d):Nothing to prove...
Here is my question:
How the one-to-one correspondence in e) actually comes out?One direction(namely given some points of $P^1$,we construct a curve)is b).The other(given a curve $f:X\rightarrow P^1$,we consider these points of $P^1$ over which $f$ has ramificaition)is a).But it's not obvious for me to see how these two procedures are actually inverse to each other...namely from $X$ by procedure in a) we get some points $p_i$,then from $p_i$ by the procedure in b) we get another curve $Y$,What's the relationship of $X$ and $Y$?
I have look up several solutions of Hartshorne about e),they all say "this follows directly from a,b,c,d" but I still don't see why.
Any reference or help is appreciated.
edit: Here is my understanding of "$f$ is uniquely determined by its branch locus up to automorphisms of $\mathbb{P}^1$".suppose $f:X\rightarrow \mathbb{P}^1$,take an affine cover $U=\text{Spec} k[x]$ of $\mathbb{P}^1$,assume ramification points of $f$ all belongs to $f^{-1}(U)=\text{Spec} A$.$A$ can be supposed to take this form $k[x,z]/(u(x,z))$,since $f$ is of degree 2,we have $u(x,z)=z^2+g(x)z+h(x)$.So the rest is to prove under some automorphism $x\rightarrow \frac{ax+b}{cx+d}$,$u$ will take the form $z^2+\prod_i (x-\alpha_i )$.Still don't know how.:(