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By the field of constructible numbers I mean the union of all finite towers of real quadratic extensions beginning with $\mathbb{Q}$. By decidable I mean the set of first order truths in this field, in the language of 0,1, + and $\times$, is recursive. Is this field either known to be decidable, or known not to be?

As of 1963 Tarski's question of whether this field is decidable was open -- so i doubt any simple adaptation of his result on real closed fields can settle this question. He conjectured that the only decidable fields were finite, real closed, or algebraically closed. See Julia Robinson, The decision problem for fields, Theory of Models (Proc. 1963 Internat. Sympos. Berkeley), North-Holland, Amsterdam, 1965, pp. 299–311. especially pages 302 and 305.

Much has gone on since 1963, and Tarski's general conjecture is well refuted, but I do not find a solution to this problem.

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Write to Videla and ask him. He surely knows the state of the art and he will be happy to answer; he is a very nice person. As I understand, the problem is open and there is a group of people (including Videla) looking into it. –  Pasten Sep 18 '13 at 14:03
I will, thanks. –  Colin McLarty Sep 18 '13 at 14:15

3 Answers 3

up vote 22 down vote accepted

According to the following paper:

Carlos R. Videla, On the constructible numbers, Proceedings of the American Mathematical Society Vol. 127, No. 3 (Mar., 1999), pp. 851-860.

The problem has remained open at least until 1999. I think the problem is still open. In the above paper the author proves that the ring of constructible algebraic integers is first-order definable in the field of constructible numbers. The author hopes that $\mathbb{Z}$ should be definable in the ring of constructible algebraic integers and therefore his result would be a partial result towards resolving the problem negatively.

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Videla politely understates. If $\mathbb{Z}$ is found definable in the ring of constructible algebraic integers then that plus his result would settle this problem negatively. –  Colin McLarty Sep 17 '13 at 14:01
Yes but this is the tradition of the mathematical society that gives the most credit to the one who takes the last step. –  shahram Sep 17 '13 at 14:18
I think it worth saying explicitly that the two results together would not just be progress, but would settle the problem. –  Colin McLarty Sep 17 '13 at 19:13

The problem of the decidability of the constructible numbers is still open. It would be very nice if it is solved ( one expects the field to be undecidable). If this is the case then Hilbert's Tenth problem would be the natural thing to attck for this field.Both results would be saying, in simple terms, "that ancient Greek mathematics was hard". On the other hand , I did prove that the field of numbers constuctible from a straightedge with a scale is undecidable. I do not know if Hilbert's tenth problem for this field is decidable. Carlos R. Videla

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Welcome to MO, Carlos! –  Todd Trimble Sep 18 '13 at 17:43

Ziegler's theorem, a special case of which is that the theory of Euclidean fields (ordered fields in which every positive has a square root) is undecidable, already shows that ancient Greek geometry was creative, not mechanical. In English, see

The constructible field is just the smallest Euclidean field (but of course we don't yet know that it is undecidable).

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Thanks for this (not the first thing you have taught me about geometry!). Your historical remark and Carlos Videla's are important perspectives. But my recent fondness for proof theory makes me hesitate some. Comments at… suggest that Euclid's practice includes no existential claims without a certain kind of definable instances, and suggest that this restriction of geometry is decidable. Probably there are important insights on both views, and more to be found by more work. –  Colin McLarty Sep 20 '13 at 13:38

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