Chevalley's valuation extension theorem and the axiom of choice

Hello,

Do we know if the axiom of choice is needed for Chevalley's valuation/place extension theorem (i.e. the theorem that states that for every valued field and a field extension, one can extend the valuation to the field extension)?

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Can somebody more expert about this retag? I'd say commutative algebra and/or algebraic number theory, but I don't really know. – darij grinberg Mar 27 '11 at 18:08

I think this more or less answers it?

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What about finite extensions? (Disclaimer: I don't know any proof of the theorem, even for finite extensions.) – darij grinberg Mar 27 '11 at 18:07
I would like to see a bit more in an answer than is provided by the answers to the linked question. They say that the answer is "yes": one needs AC (that is my guess as well), but unless I missed it no argument or reference is given. – Pete L. Clark Mar 28 '11 at 6:00
@darij: for a finite extension, there are only finitely many ways to extend the valuation, so surely AC is not needed. – Pete L. Clark Mar 28 '11 at 6:00

The MO question Artem points to is certainly relevant here, but I think it's also important to realize that the extensions of valuations studied somewhat narrowly by Krull were eventually subsumed in the much more general Chevalley extension theorem for homomorphisms. This is treated in a number of algebra books in the classical tradition, including Jacobson's Basic Algebra II (9.8) and Chapter 6 of Bourbaki's Algebre commutative. While field extensions of special types such as a simple transcendental extension might be handled more concretely, the most general extension principle formulated by Chevalley does seem to require the Axiom of Choice in the guise of Zorn's Lemma, much in the spirit of existence proofs for maximal ideals.

By the way, I think it's helpful for those asking questions to provide at least some specific references (print or online) to focus the discussion.

P.S. As Pete remarks, there is no proof offered that the Axiom of Choice is necessary here. I'm not sure how logicians view the question, but from a practical point of view I suspect most people would place the burden of proof on the other side: if you really believe AC is not needed, come up with a better proof than the existing ones. Most of us prefer constructive approaches when possible, but Chevalley's general extension theorem doesn't seem promising. The proofs I've seen all rely on Zorn's Lemma in a completely natural way, as do similar theorems in algebra. (One might even wonder whether Chevalley's theorem is equivalent to AC.)

A small question of my own: Is it true that Chevalley was a co-inventor of the label "Zorn's Lemma"? (I recall reading that somewhere, but you can't believe everything you read.)

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Choice is needed. For example, it is consistent with ZF that there exist an algebraic closure L of Q such that every absolute value on L is trivial (and Gal(L/Q) is trivial).

See: Hodges, Wilfrid. Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289--297. MR0422022

As for the origin of "Zorn's Lemma", try

Campbell, Paul J. The origin of "Zorn's lemma''. Historia Math. 5 (1978), no. 1, 77--89. MR0462876

The following is quoted from Berrick and Keating, 2000, p26: The name of the statement [Zorn's Lemma], although widely used (allegedly first by Lefschetz), has attracted the attention of historians (Campbell 1978). As a maximum principle', it was first brought to prominence, and used for algebraic purposes in Zorn 1935, apparently in ignorance of its previous usage in topology, most notably in Kuratowski 1922. Zorn attributed to Artin the realization that thelemma' is in fact equivalent to the Axiom of Choice (see Jech 1973). Zorn's contribution was to observe that it is more suited to algebraic applications like ours.} is equivalent to the Axiom of Choice, and hence independent of the axioms of set theory.

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