# algebraic geometry for specific fields means no axiom of choice?

I'm beginning to study commutative algebra and algebraic geometry, and something confuses me.

Many proofs, e.g. of the Nullstellensatz, require the axiom of choice (to get the right results about ideals, I suppose). Is there any way to avoid that axiom, if you start with an algebraically closed field that is countable? What about the complex field?

I've been looking for something on this topic on the Internet, with no luck. Thanks for any help!

• If your field is countable, you can probably get away with most things. Do note, however, that it is consistent that $\Bbb Q$ has two non-isomorphic algebraic closures (but only one of them is countable). So even that's to be taken with a pinch of salt. – Asaf Karagila Aug 3 '18 at 17:16

Many applications of choice can be removed when we restrict attention to well-orderable - or even better, countable - fields and other objects. As Asaf says, however, there are nonetheless some facts which really do require the axiom of choice. At the same time, these often have slight weakenings that (a) don't require choice and (b) do everything we really need (e.g. that there is only one countable algebraic closure of $\mathbb{Q}$ up to isomorphism).