Return to Revisions

If $k$ is an algebraically closed field of characteristic zero then the algebraic closure of $k((x))$ is $F = \cup k((x^{1/n}))$, the field of Puiseux series. In particular, the algebraic closure of $k(x)$ is contained in $F$. Now you want to look at the algebraic closure of $k(x,y)=k(x)(y)$, which is contained in the algebraic closure of $F(y)$, which in turn is contained in $\cup F((y^{1/n}))$. I think that gives you what you want.