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Let $K$ be a field, $t$ a transcendetal element over $K$ and $F'|K(t)$ an infinite Galois extension. Hence I have a tower of extension $K\subset K(t) \subset F' $. Does exist a subextension $F$ of $F'$ such that:

-trdeg$(F'|F)=1$

-$F'$ is a finite algebraic extension of $F(t)$?

In general if I have a curve $U$ over a field $K$ with function field $K(U)$ let $F'$ be the maximal galois unramified extension given by the composite of all finite separable extension $L$ of $K(U)$ such that the corresponding cover of curves $C_L\rightarrow U$ is etale. Does exist a subfield $F$ of $F'$ such that:

-trdeg$(F'|F)=1$

-$F'$ is a finite algebraic extension of $F(t)$

-$F$ is algebraically closed in $F'$?

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1 Answer 1

No. The first condition implies that $F$ is algebraic over $K$ because of the additivity of transcendental degrees. Now take $K=\mathbb C$ and $F'$ an algebraic closure of $K(t)$. If $F$ exists, it would be equal to $K$, but $F'$ is not finite over $K(t)$.

In fact the situation is even worse : for any $s\in F'$, $F'$ is infinite over $K(s)$.

Answer to the edited question. Still no. Take $K=\mathbb C$ and $U$ the affine line (parametrized by $t$) minus the origin. Then $F'=\mathbb C(t^{1/n})_{n\ge 1}$. It is infinite over $K(s)$ for any $s\in F'$. Indeed, any $s$ belongs to $\mathbb C(t^{1/n})$ for some $n$. But then $t^{1/nm}$ has degree $\ge m=[K(t^{1/nm}): K(t^{1/n})]$ over $K(s)$ for all $m\ge 1$. So $F'$ has infinite degree over $K(s)$.

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