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Let $E/F$ be a totally transcendental extension, i.e., any element in $x\in E-F$ is transcendental over $F$. Is there any field $L$ so that for some $\alpha \in L-E$, $\alpha$ is algebraic over $F$?

In other words, can we say that if $E$ is a totally transcendental extension over $F$, then any field containing $E$ is also a totally transcendental extension over $F$?

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  • $\begingroup$ If $K/F$ is a finite extension then $K(x)/F$ is not purely transcendental $\endgroup$ – reuns May 9 '17 at 0:07
  • $\begingroup$ Is there any difference between totally and purely trans. extension? $\endgroup$ – Ninja May 9 '17 at 0:19
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Take $F=Q, E=Q(X)$ and $L=E[√2]$

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Even if $E$ is totally transcendental over $F$, a field $L$ containing $E$ need not be totally transcendental over $F$. For example, let

$$F = \mathbb{Q},\;\;\;E=\mathbb{Q}(x),\;\;\;L=\mathbb{C}(x),\;\;\;\alpha=i$$

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  • $\begingroup$ Okay, I have a question about another question: math.stackexchange.com/a/2271356/201094 Then, how can we conclude the last sentence of the second answer? $\endgroup$ – Ninja May 9 '17 at 0:05
  • $\begingroup$ I don't see the issue, but maybe ask your question about that answer in that thread (I'll browse it there if you post a comment). $\endgroup$ – quasi May 9 '17 at 0:10
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Namaste May 9 '17 at 0:59
  • $\begingroup$ @amWhy: The question was "is something possible?". I showed it was possible. $\endgroup$ – quasi May 9 '17 at 1:06
  • $\begingroup$ I didn't downvote, quasi. And the message you received is a scripted message sent from the review queue, that I did not write, I think this would have made a great comment. And ditto for the other answer. $\endgroup$ – Namaste May 9 '17 at 1:11

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