Function field of genus zero with a prime of degree 2 I found this exercise on Rosen's book "Number Theory in Function Fields":
Let $F/K$ be a function field of genus zero and let $P$ be a prime of $F$ of degree $2$. Let $\{1, x, y\}$ be a basis for the Riemann-Roch space $\mathcal L(P)$ (you can show that the dimension is exactly 3 by using Riemann-Roch theorem). Show that $F=K(x,y)$.
My idea was to use the fact that $[F:K(x)]=2$ (I know this by theory), but then I can't prove that $y$ is not contained in $K(x)$. Is this the right approach? How would you prove it?
 A: I don't know if it's still relevant, but I think I have a solution. I hope I did not make any mistake, and that it will be of any help to you.
Let's assume towards contradiction that $F(x)=F(x,y)=F(y)$ (By symmetry, if one of the equalities is not true, we finish by your argument). By the properties of the p-valuation $v_p$, we know that $\{1,x,y,x^2,y^2,xy\}\in L(2p)$. By Riemman-Roch the dimension of $L(2p)$ is 5. So we have a linear dependency,
$$a_0+a_1x+a_2y+a_3x^2+a_4y^2+a_5xy=0$$
The ring $F[T,U]$ is a UFD. So if we denote $f(T,U)=a_0+a_1T+a_2U+a_3T^2 +a_4U^2+a_5TU$, and if $f(T,U)$ is reducible, it must be factored to linear factors. But then by the previous equation we get a non-trivial linear dependency on $1,x,y$ (over $F$) which is impossible. So, $f$ is irreducible.
Now look at the polynomial $a_0+a_1x+a_3x^2+(a_2+a_5x)T+a_4T^2\in F(x)[T]$. If $a_4\neq 0$, it is irreducible, since it is a primitive, irreducible polynoimal over $F[x][T]$. If $a_4=0$, we may assume WLOG that $a_2+a_5x$ does not divide $a_0+a_1x+a_3x^2$. Other wise, we could devide the first equation to ge an equation of the form $b_0+b_1x+y=0$, which is impossible by the linear independence of $1,x,y$. Hence The polynomial above is still primitive and so is the minimal polynomial of $y$ over $F(x)$. 
We split to 2 cases. If $a_3=0$, we may write:
$$a_0+a_1x+a_2y+a_5xy=0$$
As $a_5\neq 0$ (by linear independence of $1,x,y$) This gives us-
$$y=\frac{-a_1x-a_0}{a_2+a_5x}\Rightarrow v_p(y)=v_p(-a_1x-a_0)-v_p(a_2+a_5x)$$
As the $p$ valuation of $1,x$ are different, we know that $$v_p(\alpha+\beta x)=min(v_p(\alpha),v_p(\beta x))=min(0,-1)=-1$$
If $\alpha\neq 0$. So, we get $v_p(y)\geq -1+1=0$ - a contradiction. 
So we may assume $a_3\neq0$. In this case, by the symmetry between $x$ and $y$ - we finish.
[Edit: I now noticed that I am using a different notation than you, taking $F$ to be the ground field.]
