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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.

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The fact that every ideal is contained in a maximal ideal requires the axiom of choice. –  David Speyer Dec 8 '11 at 18:03
    
@David Oops! Didn't look far, did I? –  M Turgeon Dec 8 '11 at 18:21
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Huh?${}{}{}{}{}{}$ –  Asaf Karagila Jul 4 '12 at 21:36
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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 5 '12 at 3:38
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@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 5 '12 at 6:08

1 Answer 1

up vote 17 down vote accepted

"...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.

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3  
Beat me to it by a few seconds! See consequences.emich.edu/conseq.htm and mathoverflow.net/questions/46566/… for some details. –  Qiaochu Yuan Dec 8 '11 at 18:07
    
@QiaochuYuan: Thanks for the links! –  Arturo Magidin Dec 8 '11 at 18:10
    
Thanks to you both! –  M Turgeon Dec 8 '11 at 18:25

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