Let $\tau:F \to \overline{F}$ be a field embedding.

Then is $\overline{F}/\tau(F)$ algebraic extension?

I don't think so but I cannot find a counterexample.

Would you let me know a counterexample?

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    $\begingroup$ You said "also". Are you assuming something (which you haven't mentioned) is algebraic over something else? Consider any field $k$, adjoin a variable $X$ and consider $k(X)/k$, and the inclusion $k\to k(X)$? If you're asking if non-algebraic extensions exist, they certainly do. $\endgroup$
    – Pedro
    Jul 12, 2016 at 4:54
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    $\begingroup$ @PedroTamaroff: I think the question is saying: "$\bar{F}/F$ is an algebraic extension. Given an embedding $\tau \colon F \to \bar{F}$, is $\bar{F}/\tau(F)$ an algebraic extension as well?" $\endgroup$ Jul 12, 2016 at 4:56
  • $\begingroup$ Is $\bar F$ the algebraic closure of $F$? If that's the case, the result is very trivial. $\endgroup$
    – JasonM
    Jul 12, 2016 at 5:24

1 Answer 1


It is a bit unclear what is asked, but judging from the comments of others the following interesting interpretation comes to mind. We start with a field $F$ and its algebraic closure $\overline{F}$, and then we wonder whether $\overline{F}/\tau(F)$ is algebraic for all field homomorphisms $\tau:F\to\overline{F}$.

The answer to that question is 'No'.

Let $F=\Bbb{Q}(x_0,x_1,\ldots)$ be a purely transcendental extension of the rationals of a countably infinite transcendence degree. Let us define $\tau:F\to\overline{F}$ by declaring $\tau(x_i)=x_{i+1}$ and extending that to a homomorphism of fields in the obvious way. Then the element $x_0\in\overline{F}$ will be transcendental over $\tau(F)$.

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    $\begingroup$ Beautiful, Jyrki. I was suspicious the answer was indeed no, but couldn't come up with an example myself. Thanks for posting! $\endgroup$ Jul 12, 2016 at 5:37
  • $\begingroup$ @Jyrki, Thank you very much! Your answer is what I expected! $\endgroup$
    – user29422
    Jul 12, 2016 at 5:44

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