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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 ?

flag	asked Sep 12 at 20:55 Feldmann Denis 1,768●2●15

Andreas Blass · Accepted Answer · 2012-09-12 21:17:33Z UTC

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.

Igor Rivin · Answer 2 · 2012-09-12 21:21:28Z UTC

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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 at 21:21

Igor Rivin
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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 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 at 1:29

Is Monsky’s theorem depending on axiom of choice ?

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