The extension of the 2-adic valuation to the reals used in the usual proof clearly uses AC. But is this really necessary? After all, given an equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken?
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asked Sep 12, 2012 at 20:55
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4 Answers
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No choice is needed. If, in a choiceless universe, there were a counterexample, then that counterexample amounts to finitely many real numbers (the coordinates of the relevant points). It would still be a counterexample in the sub-universe of sets constructible (in Gödel's sense) from those finitely many reals. But that sub-universe satisfies the axiom of choice, so your favorite ZFC proof of the theorem applies there.
answered Sep 12, 2012 at 21:17
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1$\begingroup$ Am I right that this only shows $\def\zf{\text{ZF}}$$\zf \vdash φ$ where $φ$ is a sentence stating Monsky's theorem? If so, then am I correct that it is weaker than the other proof, because it's possible that ZF is $Σ_1$-unsound in which case you can prove that ZF proves $φ$ although there is in fact no proof in ZF of $φ$? The other proof however gives us an explicit way to write down a proof of Monsky's theorem over ZF, which avoids the need to assume that ZF is $Σ_1$-sound. $\endgroup$ Commented Jul 28, 2016 at 11:32
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2$\begingroup$ @user21820 No, this argument, combined with the proofs in ZF of the relativizations to $L[x]$ of the ZFC axioms, provides a way of (constructively) converting any ZFC proof of $\varphi$ into a ZF proof of $\varphi$. Since a ZFC proof is known, we could (tediously) produce an explicit ZF proof. $\endgroup$ Commented Jul 28, 2016 at 17:48
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I was recently looking over Monsky's Theorem as supplementary material for my course notes on local fields, and I noticed that his original article (available here) ends by addressing your question:
The above proof is not so wildly nonconstructive as it first appears. For the entire argument is carried out in the field generated by the coordinates of the vertices. So it is only necessary to extend our ultranorm from $\mathbb{Q}$ to this finitely generated field, not to the entire field of real numbers.
It is easy to see that extending a rank $1$ valuation from a field $K$ to any monogenic extension $K(t)$ does not use the axiom of choice: if $t$ is algebraic over $K$ the set of extensions is finite, nonempty and explicitly in bijection with $\operatorname{Spec} \hat{K} \otimes_K K(t)$ (and even without AC a finite-dimensional $K$-algebra must have a maximal ideal!); if $t$ is transcendental over $K$, we may endow $K(t)$ with the Gauss norm, determined on $K[t]$ by $|a_n t^n + \ldots + a_0| = \max_i |a_i|$ and extended to $K(t)$ by multiplicativity.
Otherwise put: whereas Andreas Blass's nice answer explains why any proof of this result yields an AC-less proof, my answer mentions that Monsky's proof does not really use AC, as pointed out by Monsky.
answered Sep 6, 2013 at 21:38
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This paper (Projective Colorings, by Hales and Straus) seems to imply that the Axiom of Choice is necessary for closely related results.
answered Sep 12, 2012 at 21:21
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1$\begingroup$ The use of the axiom of choice in the Hales-Strauss paper is needed to get non-Archimedean valuations on the whole real field. But in the proof of Monsky's theorem, they use the valuation only for the coordinates of the points involved in an alleged counterexample to Monsky's theorem. So, any particular alleged counterexample can be refuted by a non-Archimedean valuation on a much smaller field, in fact a countable subfield of the reals. And the existence of such valuations doesn't need choice. $\endgroup$ Commented Sep 13, 2012 at 0:18
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$\begingroup$ Thanks! This is pretty much as the OP had conjectured, but I guess my point was that we were skating pretty close to the AC... $\endgroup$ Commented Sep 13, 2012 at 1:29
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In the proof given in the book "Proofs from the book" by Martin Aigner and Gunter Ziegler , 2-adic norm is not specifically used.
It proves it using an arbitrary non-archimedean norm such that v(1/2) > 1. And proves that such a validation exists for R .
Thought it also mentions that 2- adic norm can be extended to reals but it is not "a standard algebra fare" although I don't know what that means.
Here is the link for the proof https://books.google.co.in/books?id=2iI9BAAAQBAJ&pg=PA156&lpg=PA156&dq=monsky+theorem+real+numbers&source=bl&ots=ZnUF1Jv8yF&sig=jLjRaeJYmxtQaYCayy3IkqjRUg0&hl=en&sa=X&ved=0CEoQ6AEwB2oVChMI-9SMv9XnxwIVEnGOCh245gwy#v=onepage&q=monsky%20theorem%20real%20numbers&f=false
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3$\begingroup$ How does this answer the question about the axiom of choice in the proof? $\endgroup$– Asaf Karagila ♦Commented Sep 8, 2015 at 18:14
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$\begingroup$ I tried to imply that since it proves the existence of non -archimedian validation for R without axiom of choice it should be able to extend 2-adic to R without it too. Am I mistaken in this line of thinking? $\endgroup$ Commented Sep 19, 2015 at 21:22