Is Monsky's theorem dependent on the axiom of choice? 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?
 A: 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.
A: 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.
A: This paper (Projective Colorings, by Hales and Straus) seems to imply that the Axiom of Choice is necessary for closely related results.
A: 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
