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The extension of the 2-adic valuation to the reals used in the usual proof uses clearly AC. But is this really necessary ? After all, given a 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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up vote 13 down vote accepted

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.

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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.

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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.

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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. –  Andreas Blass Sep 13 '12 at 0:18
    
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... –  Igor Rivin Sep 13 '12 at 1:29
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