# Global choice and algebraic closures

December 28, 2011 2 Comments

It was pointed out to me today that I glazed over a set-theoretic point in my proof that every field has an algebraic closure. We appealed to Zorn’s lemma, which says:

Given a partially ordered set , if every chain has an upper bound, then has a maximal element.

An assumption we made here that seems innocuous is that is a set. If it is a proper class (class and not a set), we may run into problems. So what if the collection of all field extensions of a given field is a proper class? Then we can’t use Zorn’s Lemma.

I think, though haven’t attempted to prove it, that this collection of field extensions is really a set, meaning we’re safe in applying Zorn’s lemma. But it’s an interesting diversion to see what happens if we try to prove Zorn’s lemma for classes.

It’s well known that Zorn’s lemma is equivalent to the axiom of choice, so it would be nice to find an “axiom of choice for classes.” Such a thing exists, and it’s called the “axiom of global choice.” It says the same thing about classes that the axiom of choice says about sets. So if you accept the axiom of global choice, we just need to prove a “global Zorn’s lemma.”

Unfortunately, the standard proof of Zorn’s lemma goes something like this: Assume it’s false. Then use the axiom of choice to construct (via transfinite induction) larger and larger elements. These form a well-ordering. We can keep going and we’ll eventually get to a well-ordering that has too many elements to be contained in the set .

The problem is that if we replace the axiom of choice with the axiom of global choice, we’ll be unable to make this well-ordering bigger than our “poclass” .

If you want more information on this sort of set theory stuff, you should read Zach Norwood’s posts (this is a great blog to which you should definitely subscribe):

On an unrelated note, coming soon will be a series of posts on some generalized abstract nonsense.

Awesome! Can’t wait for category theory! And thanks for the links to Zach’s posts, they are indeed worth reading. 🙂

The class of all extensions is, in fact, a proper class. If it weren’t, then your proof would be correct, but in fact, there is no such thing as a maximal field. Any field is a proper subfield of the field of rational functions over that field.