# Talk:Field has Algebraic Closure

## Issues

Thinking a bit more about it, it is clear that the collection of all proper extensions of any non-algebraically-closed field is a proper class. If $(F,+,\cdot)$ is a field and $(G,+',\cdot')$ is an extension of it, let $x \in G \setminus F$. Then $x$ can be replaced by any set in $\mathbb U \setminus G$ to form another (isomorphic) extension. This particular aspect of the problem could be handled in well-founded set theory by selecting only extensions that are of minimal rank within the isomorphism class, but I think it will still be difficult, if not impossible, to show that $\mathcal F$ is a set. However, it appears that our definition of algebraic closure is wrong/incomplete. Specifically, it appears that the algebraic closure of a field must be algebraic over that field. That would seem to have quite a bit of potential to bring things down to a manageable size. I don't know terribly much about algebra, but it appears that for an extension to be algebraic over a field, each element of the extension must be a root of a non-zero polynomial over the field. The collection of polynomials over any given field certainly is a set. I imagine, then, without knowing for sure, that there is probably some way of representing all potential solutions to each polynomial (e.g., as a disjoint sum of finite ordinals). Anyway, this construction and others, all much more involved than what we currently have, are presumably correct. So the theorem is right, but the proof is probably not salvageable. --Dfeuer (talk) 15:26, 13 May 2013 (UTC)

This MathOverflow discussion links to some papers on the topic. It is known that BPI implies that every field has an algebraic closure. This StackExchange question outlines what is apparently the classic proof using Zorn's lemma (but the comments below suggest some preliminary work is needed to avoid the very same issue I raised above). --Dfeuer (talk) 16:03, 13 May 2013 (UTC)

I think you're right: any field has a prime subfield which is $\mathbb F_p$ or $\Q$; so the (proper) class of all fields is the class of all extensions of $\mathbb F_p$ and of $\Q$ (this argument is correct provided we define a field extension to be a morphism $E \to F$ rather than an inclusion $E \subseteq F$; in the latter case I guess it's still a proper class, but the stated argument only gives an equivalence of categories). Thus it cannot be the case for every field that the class of all extensions is a set. --Linus44 (talk) 17:02, 13 May 2013 (UTC)

As I showed above, proper inclusion is quite a big enough problem, because if $G$ properly includes $F$, then even the class of all fields that properly include $F$, are isomorphic to $G$, and differ from $G$ by only one element is a proper class. --Dfeuer (talk) 17:34, 13 May 2013 (UTC)
Unfortunately, I don't yet know enough algebra to be able to put up a valid proof. Do you? --Dfeuer (talk) 17:36, 13 May 2013 (UTC)

Agreed, your argument works in this case as well.

I have a proof in some notes I took for during course on Galois theory; if I recall correctly following the construction you mention, but my memory of it is vague. If it's not too much work I'll write it up. --Linus44 (talk) 18:30, 13 May 2013 (UTC)

For the moment, it is too much work. --Linus44 (talk) 18:51, 13 May 2013 (UTC)