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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
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
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. – Andrés Caicedo Dec 30 '13 at 17:41
up vote 10 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 '14 at 8:10
There is, if you work in the category of sets. – Asaf Karagila Jun 11 '14 at 8:14
Ah, you're just using $F[x]$ to have a concrete set with the right size. – blue Jun 11 '14 at 8:18
Yeah, that's the idea. – Asaf Karagila Jun 11 '14 at 8:21

The other answers rescue your quoted proof of the existence of algebraic closure by observing that there is an explicit bound you can place on the cardinality of the algebraic closure. Here is an alternative proof that side-steps the set-theoretic issue altogether:

Let $S$ be the set of all irreducible polynomials with coefficients in $K$, and let $$R := K\left[\{x_\alpha\}_{\alpha \in S}\right]/\left(\{f(x_\alpha)\}_{\alpha \in S}\right).$$ Then $R$ has a maximal ideal, so we can define $F := R/\mathfrak m$ for $\mathfrak m$ such a maximal ideal, which makes $F$ into an algebraic extension of $K$ in which every polynomial in $S$ has a root.

At this point, it is true that $F$ is algebraically closed, but not easy to prove. You can avoid this additional piece of machinery by simply defining $F_1$ to be the field obtained above, and iteratively defining $F_2, F_3, \ldots$ by the same procedure. A priori, each $f \in S$ can readily be shown to split in $F_i$ for $i \ge \deg f$, and so $F := \displaystyle\bigcup_i F_i$ is an algebraic closure for $K$.

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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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To answer your other question, of "how to do concrete everyday mathematics, avoiding set theory, without getting burnt":

In NBG set theory, a class (where "class" means "collection of sets") is a set if and only if it is "small"—which is to say, strictly smaller than the class of all sets. In addition, every theorem of NBG that doesn't use the word "class" is a theorem of ZFC.

So, here's how to avoid getting burnt. Think of any class. (A class can be defined by any predicate that only talks about sets, not classes.) If you can show that your class is small, that means it's a set. If you make sure all of your sets are created this way, you end up with a theorem of ZFC.

How do you know that your class is small? These axioms should do the trick:

  • The class of all integers is small.
  • Given any small class, the class of all of its subclasses is a small class.
  • Given any small class, the union of its elements is a small class.

Intuition should give you everything else you need.

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what does "strictly smaller than the class of all sets" mean ? isn't the union of the underlying sets of the elements of the collection of algebraic extensions of $K$ the same as the class of all sets ? – mercio Aug 12 '14 at 15:49
Answer to first question: "A is strictly smaller than B" means that there is no surjective function from A to B. Answer to second question: yes, it is; this shows that the collection of algebraic extensions of $K$ is not a small class. – Tanner Swett Aug 12 '14 at 16:08
The last comment involves global choice. Otherwise there is no surjection from the class of ordinals onto the class of all sets. It's better defining "small" here as no surjection onto the class of ordinals, and then even if the axiom of choice (for sets) fails badly, you still retain the same meaning of sets are small classes. – Asaf Karagila May 7 '15 at 6:38

Another way to legalize your reasoning - Grothendieck universe. By the Grothendieck axiom there exists universe $\mathcal{U}$ which contains $K$. Let $A=\{F:F-\text{algebraic extension of}\ K\ \text{and}\ F\in\mathcal{U}\}$. Then all the problems disappear. Here we use the fact that universe is closed under many constructions, including quotient ring.

By the way, in the same spirit it is possible also to get various free constructions, such as free groups, free abelian groups, polynomial rings,...

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The answer equivalent to "if you want to kill a mosquito turn the sun into a black hole"... – Asaf Karagila Mar 20 at 16:08
@ Asaf Karagila If "universe" is a black hole for you, then you are right. In your universe ) – Alex W Mar 20 at 17:58
I mean that in the sense of that it is far of an overkill solution, for something which is only mildly annoying to deal with. – Asaf Karagila Mar 20 at 18:01
@Asaf Karagila Who knows ) But definitely universes are natural in some sense, and they relieve us from the boring arguments with bounds for cardinalities and embeddings in sufficiently large set. – Alex W Apr 1 at 13:23
Your edit doesn't make any sense. Why do you need universes for constructing a free group? $\Bbb Z$ is a fixed set. Therefore given any $S$, the free group generated by $S$ is a very concrete and a pretty specific set. Using universes here is the same as using a nuclear bomb to take a shower: it makes no sense. – Asaf Karagila Apr 1 at 13:30

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