Show the algebraic closure $\bar{\mathbb{Q}}$ is not even finitely generated over the field $\mathbb{Q}$

I'm not sure how to go about this..


closed as off-topic by Servaes, Vinyl_cape_jawa, max_zorn, Leucippus, Parcly Taxel Mar 1 at 6:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Vinyl_cape_jawa, max_zorn, Leucippus, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.


Hint: Show that for any natural $n$, there is an algebraic extension of $\Bbb Q$ of degree (at least) $n$. And any finitely generated algebraic extension has some finite degree.

  • $\begingroup$ Could I say that since [Q bar : Q] = infinity, after what you suggested, would be a contradiction and thus proving the premise? $\endgroup$ – Christian Feb 28 at 23:18
  • $\begingroup$ @Christian No contradiction at all. Just proving $[\overline{\Bbb Q}:\Bbb Q]\geq n$ for all $n$. That's all we need. $\endgroup$ – Arthur Mar 1 at 0:56
  • $\begingroup$ Thank you so much! $\endgroup$ – Christian Mar 1 at 20:50

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