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Having read the classical proof of the existence of an Algebraic Closure (originally due to Artin), I wondered what is wrong with the following simplification (it must be wrong, otherwise why would we bother with the extra complications in Artin's proof?):

Theorem: Let $K$ be a field, then $K$ has an algebraic closure $\bar{K}$ (i.e an algebraic extension that is algebraically closed).

"Proof": Define $A=\{ F \supset K | F \text{ is an algebraic extension of } K\}$ and inherit this with the usual partial order of inclusion. One can check that Zorn's lemma applies (union of a nested chain of algebraic extensions is itself algebraic). Thus take $\overline{K}$ to be a maximal element. It must be algebraically closed for otherwise there is an irreducible polynomial with root in some strictly bigger field. $\blacksquare$

Now here is what I suspect is false about this proof: The definition of $A$ smells like your usual set theory paradoxes like Russell's paradox. In fact one could just as well use the same technique to prove that there exists a "largest set" which of course there does not. I am however under the impression that "most" working mathematicians ignore set theory foundations and just "do" mathematics, so is there a safe way of doing this (i.e: do "concrete everyday mathematics" by avoiding set theory) without getting burnt?

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I don't see why Zorn's lemma applies. –  Igor Rivin Dec 30 '13 at 3:33
    
@Igor OP says why in the very same sentence: just take a union of any totally ordered collection of algebraic extension and again you have an algebraic extension which is an upper bound of the given collection. –  anon Dec 30 '13 at 3:34
    
@anon yes, I guess that works. –  Igor Rivin Dec 30 '13 at 3:38
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It seems to me that set-theoretic issues are avoided if, instead of the nebulous "all" algebraic extensions, you just consider those extensions which are contained in some fixed but sufficiently roomy set, e.g. $\mathcal P\mathcal P\mathcal P(K\cup\omega)))$ or something like that. –  bof Dec 30 '13 at 3:46
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As pointed out in the comments, there is an issue (this is a proper class), but it can be easily dealt with. Some authors make a big deal out of this (D.J.H. Garling, for instance), but the fuss only reveals that they do not feel too comfortable with the underlying set theoretic machinery. –  Andres Caicedo Dec 30 '13 at 17:41

3 Answers 3

up vote 6 down vote accepted

In fact, there is a slightly goosed version (by Jelonek) which avoids the set-theoretic issues. Poles know from set theory...

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You are right. $A$ is a proper class. The reason is simple, by considering all possible fields which are algebraic extensions we immediately have a proper class of sets.

However one can easily observe that if $F$ is a field, then there is a map from $F[x]$ onto any algebraic extension, therefore it suffices to consider algebraic extensions whose underlying set is a partition of $F[x]$.

In either case, one can show that despite the fact that $A$ is a proper class, it is "locally a set", in the sense that below each field there is only set-many fields; and that every chain has size no larger than $|F|+\aleph_0$.

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There is no onto map from $F[x]$ to an infinite algebraic extension of $F$. –  blue Jun 11 at 8:10
    
There is, if you work in the category of sets. –  Asaf Karagila Jun 11 at 8:14
    
Ah, you're just using $F[x]$ to have a concrete set with the right size. –  blue Jun 11 at 8:18
    
Yeah, that's the idea. –  Asaf Karagila Jun 11 at 8:21

This is the first proof I learnt of the existence of algebraic closures, from Fraleigh's book (an edition from the mid-to-late 80s). As you note, you have to be a little bit careful with set-theoretic issues, but that is not so serious.

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