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I'm interested in investigating the field extensions of algebraically closed fields. Suppose we have a field extension $F \subseteq K$ where both $F$ and $K$ are algebraically closed. From the definition, I think we can show that every element in the extension $K/F$ must be transcendental, however, I don't think it's a purely transcendental extension.

I've seen examples like this

Why is $\mathbb{Q}(t,\sqrt{t^3-t})$ not a purely transcendental extension of $\mathbb{Q}$?

from Dummit & Foote, but I don't know how to show it here. I've started by assuming that it is a purely transcendental extension and seeking a contradiction, but I don't know where the contradiction comes from.

For example, I thought I could start something like this: If $K/F$ is a purely transcendental extension, then we can write $K = F(X)$ for some transcendence base, $X$.

Any help is appreciated. TIA

lia.riley
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  • If the transcendental basis is countable (if not it should be the same with a transfinite recursion) then $K = \bigcup_{n\ge 0} E_n$ where $E_0=F,E_{n+1}= \overline{E_n(x_{n+1})}$, and it suffices to understand the extension $\overline{k(t)}/k(t)/k$. – reuns Apr 09 '20 at 02:20

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A purely transcendental extension has the form $K=F((x_\alpha)_{\alpha\in A})$, that is it is generated by a set of algebraically independent elements $x_\alpha$ indexed by a set $A$. Unless $A$ is empty (i.e., $K=F$) $K$ cannot be algebraically closed, since for any $\alpha\in A$, $x_\alpha$ is not a square in $K$.

Angina Seng
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