When working with fields, it's a usual method to work on an algebraic closure of a field to obtain results about that field. In general (i. e. unless you're explicitly considering "well-behaved" fields like $\Bbb R$ or $ \mathbb{F}_q$), the existence of this algebraic closure is only provided by the Lemma of Zorn. Thus, proofs in field theory using the algebraic closure are not valid if you don't believe in the axiom of choice.

On the other hand, algebraic closures often are used to simplify the proof, but you could also get along without them. But are there any important cases where one is dependent on the existence of algebraic closures? In other words: Do you know theorems involving fields, which can only be proved if you use the existence of algebraic closures?

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    $\begingroup$ Do you have examples for theorems which get simpler to prove over the algebraic closure (and are not applied to well-behaved fields to begin with)? $\endgroup$ – Asaf Karagila Dec 7 '12 at 15:08
  • $\begingroup$ Take for example this course in differential galois theory: www4.ncsu.edu/~singer/papers/Crespo_Hajto.pdf. On page 17 in the last paragraph regarding the proof of proposition, they tensor R with $\bar{K}$, enabling them to assume WLOG that K is algebraically closed. Unfortunately, there is no simpler example coming to my mind. $\endgroup$ – Dominik Dec 8 '12 at 18:58
  • $\begingroup$ I have to admit that I have a hard time answering you on that. I do know that it is still open whether or not the existence of an algebraic closure implies some other more familiar choice principles (in particular the ultrafilter lemma which is sufficient to prove it), and so it's still quite difficult to answer your question exactly. It is known that you can't prove the existence of a unique algebraic closure to the rationals without some choice (and so essentially to any other field), but it's not a very useful answer, is it? $\endgroup$ – Asaf Karagila Dec 8 '12 at 19:32

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