computing the full constant field of an algebraic function field

Question

Let $K$ be a field such that char$(K) \neq 2$. Let $F=K(x,y)$ be an algebraic function field of one variable $x$ where $$y^2 = f(x) \in K[x].$$ We want to compute the full constant field of $F$ (i.e. algebraic closure of $K$ in $F$). Is there a way to do that?

Answer 1

There is a standard theorem:

Theorem. If $G(x,y)$ is an irreducible polynomial in $K[x,y]$, and $F=K(x,y)$ be algebraic extension of $K(x)$ defined by $G(x,y)=0$, Then $F\cap \bar{K} = K$ iff $G$ is absolutely irreducible (i.e. irreducible in $\bar{K}[x,y]$).

In the case $G(x,y)=y^2-f(x)$ there are two cases:

1. $G$ is absolutely irreducible.
2. $f(x) = \alpha h(x)^2$ with $h(x)\in K[x], \alpha\in K\backslash K^2$. Then $$\left(\frac{y}{h(x)}\right)^2 = \alpha,$$ and $F=K[\sqrt{\alpha}](x)$ so $F\cap \bar{K} = K[\sqrt{\alpha}]$.