Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


-$F'$ 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:


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

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

share|cite|improve this question

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 $K(s)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.