12

It is well-known that, assuming the axiom of the choice (in the form of Zorn's lemma), one can prove that any field $F$ has an algebraic closure. One proof roughly goes as follows: consider the partially-ordered set of algebraic extensions $K/F$, ordered by inclusion; show that it satisfies the hypotheses of Zorn's lemma; then Zorn's lemma implies the existence of a maximal element; show that this maximal element is algebraically closed.

Now, in my Algebraic Number theory class, the professor gave a "constructive" proof of this fact: let $S$ be the set consisting of all non constant polynomials in $F[X]$, and construct the ring $R=F[X_f:f\in S]$ (that is, a polynomial ring with one variable for each element of $S$); let $A_0$ be the ideal of $R$ generated by polynomial $\{f(X_f):f\in S\}$; show that it is a proper ideal, and so it is contained in a maximal ideal $A$; the quotient $R/A$ is thus a field, which contains $F$ as a subfield; iterate, and take the union of all these fields; then, this union is the algebraic closure of $F$. He also mentionned (if I recall correctly) that one can prove that after only one iteration, you get the algebraic closure. Now, I put quotation marks around "constructive", because I suspect that one still needs the axiom of choice at some point to prove that, indeed, you have the algebraic closure (although I haven't checked the details).

So, here's my question:

Is it the case that this proof is constructive?

Another question:

Can one prove the existence of an algebraic closure within ZF (without the axiom of choice)?

Thanks in advance.

M Turgeon
  • 10,419
  • 19
    The fact that every ideal is contained in a maximal ideal requires the axiom of choice. – David E Speyer Dec 08 '11 at 18:03
  • @David Oops! Didn't look far, did I? – M Turgeon Dec 08 '11 at 18:21
  • But the "real" question, is the second one, thus the title – M Turgeon Dec 08 '11 at 18:22
  • If your professor thinks that this is a constructive proof which does not use the axiom of choice then something is a little bit wrong here. It is true that many people are unaware to the use they make of the axiom of choice almost all the time; however Zorn's lemma is boldly used to find that maximal ideal in every course... – Asaf Karagila Dec 08 '11 at 22:01
  • The problem is worse than the previous answers suggest: at least one edition of Lang gives an incomplete 'proof' of the theorem. The iteration described in the question requires not only that every ideal in the construction be contained in a maximal ideal, but also that we have a way of choosing among the possible maximal ideals. Unless we have a choice principle for proper classes, this requires a preliminary construction so that we can apply a choice principle to a set. –  Jul 04 '12 at 21:33
  • Lang omits the preliminary construction, apparently because he does not notice when he is using the general principle of choice. The application of Zorn's Lemma to the 'set' of algebraic extensions may require similar precautions. On the other hand, important special cases of Artin's theorem can be proved without choice principles. This may follow from the work of Banaschewski: if I can't find the result in the literature I may write it up myself. –  Jul 04 '12 at 21:33
  • 1
    Huh?${}{}{}{}{}{}$ – Asaf Karagila Jul 04 '12 at 21:36
  • "... after only one iteration, you get the algebraic closure." This is Lemma 1 of http://sierra.nmsu.edu/morandi/notes/algebraicclosure.pdf . The approach (not clear whether it's original) removes the set-theoretic problems which I mentioned, but requires a lot more algebra. Morandi assumes that the reader already knows finite-dimensional Galois theory. –  Jul 05 '12 at 03:07
  • @Asaf: "Huh?" Is this a question about my previous answer? If you write out the iterative "construction" you will find that you must choose a maximal ideal at each step. Since there are infinitely many steps, you need a choice principle. If you look carefully at the usual statements of the Axiom of Choice and General Principle of Choice then you will see that they do not apply without a preliminary construction. –  Jul 05 '12 at 03:07
  • 1
    You should be able to comment on your own answers, to reply to someone else's comments, instead of posting a new answer. I'll flag this so moderators can decide whether to turn them into comments. – Arturo Magidin Jul 05 '12 at 03:38
  • 2
    @Jorgen: I believe I looked enough at the statement of The Axiom of Choice. In fact I looked a lot at equivalent statements, and often I look a lot at weak choice principles. I still have no idea what you are trying to say. – Asaf Karagila Jul 05 '12 at 06:08
  • @Asaf: Perhaps it's the fact that if you consider "all algebraic extensions" of a field $F$, they form a proper class, not a set. Usually, one first bounds the size of an algebraic extension, and then only considers those algebraic extensions whose underlying sets are subsets of a "suitably large" fixed set. I don't have my copy of Lang to see whether this is an issue, but perhaps this is what the latter part of the comment may refer to. The "iterate" part of the argument above probably needs some form of dependent choice, in addition to the use of AC in the maximal ideal argument. – Arturo Magidin Jul 05 '12 at 18:29
  • @Arturo: Well, I suppose this is true. Scott's trick does wonders for these problems, and I'm usually applying it without noticing. It is also not hard to note that an algebraic closure of $F$ must have cardinality of at most $|F|\cdot\aleph_0$ so we can assume some particular set is the underlying set. I can understand the iteration part, but it is an overkill in choice terms. It is just convenient when the full axiom of choice is present, but this is not the "accurate" argument. Using the Ultrafilter lemma you can just swat all those choices into one extension of a filter to an ultrafilter!! – Asaf Karagila Jul 05 '12 at 19:53
  • @Asaf: You won't get arguments from me, of course. I'll note that we were getting a second-hand report of the proof, and that even that report mentions that it is not necessary to "iterate" (that once you take the quotient by a maximal ideal, you have the algebraic closure). This follows from the theorem that says that if $K$ is a field, algebraic over $F$, and every nonconstant polynomial in $F[x]$ has at least one root in $K$, then $K$ is algebraically closed. So you don't even need to worry about iterations, just about the existence of that maximal ideal/ultrafilter. – Arturo Magidin Jul 06 '12 at 00:16
  • @Asaf Isn't the OP's first argument wrong? I want a verification. The collection of all extensions of $F$ Is not a set so one cannot apply zorn's lemma. Am I correct? – Rubertos Mar 16 '15 at 02:44
  • @Rubertos You can show that there exists a set that contains all algebraic extensions up to isomorphism – M Turgeon Mar 30 '15 at 14:26

1 Answers1

24

"...is a proper ideal, and so is contained in a maximal ideal of $A$."

The fact that proper ideals are contained in maximal ideals of rings with unity is actually equivalent to the Axiom of Choice, so this isn't really avoiding AC, it's just hiding it.

There are models of ZF in which there are fields with no algebraic closure, so it's known that you need at least some Choice to prove that every field has an algebraic closure.

There is an article by Bernhard Banaschewski, Algebraic closure without choice, Z. Math. Logik Grundlag. Math. 38 (1992), no. 4, 383-385, which says that you can prove that every field has an algebraic closure if you only assume the Boolean Ultrafilter Theorem, which is strictly weaker than AC.

By the way, the argument given is attributed in Lang's Algebra to Artin.

Arturo Magidin
  • 398,050
  • 4
    Beat me to it by a few seconds! See http://consequences.emich.edu/conseq.htm and http://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent for some details. – Qiaochu Yuan Dec 08 '11 at 18:07
  • 1
    @QiaochuYuan: Thanks for the links! – Arturo Magidin Dec 08 '11 at 18:10