3
$\begingroup$

I have heard that an algebraic closure of $\mathbb{Q}$ can be constructed without Zorn's lemma and so can an algebraic closure of a finite field $\mathbb{F}_p$. What about $\mathbb{F}_p(T)$? Do there exist fields for which it is not possible to conclude that that there is an algebraic closure without Zorn's lemma?

$\endgroup$
7
$\begingroup$

On a technical note, the existence (and uniqueness) of algebraic closures follow from a weaker assumption than Zorn's lemma.

But we can treat this question as asking "without any appeal to the axiom of choice". Countable fields, and in fact any well-ordered field has a canonical algebraic closure which we can construct.

It may be, however, that there are two non-isomorphic algebraic closures.

Related:

Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?

$\endgroup$
  • $\begingroup$ Interesting! I am confused by something you say though... how is it that that the axiom of choice can be used to prove existence and uniqueness of an algebraic closure of any field, but in your final sentence you say algebraic closures may be non-isomorphic. Doesn't uniqueness mean 'up to isomorphism'? I don't understand this point. If the final sentence were true, it would mean that the axiom of choice proof leads to a contradiction... $\endgroup$ – JessicaB Mar 12 '13 at 16:17
  • 1
    $\begingroup$ Jessica, the answer is in the context where the axiom of choice fails. It might be the case where $\Bbb Q$ has two non-isomorphic algebraic closures. If we have some choice then we can prove this never happens, but if we don't have choice at all it might be that this is what happens in the universe. $\endgroup$ – Asaf Karagila Mar 12 '13 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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