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is the question whether a polynomial is non-negative on some semi-algebraic set (equivalently, is it in the cone denerated by some polynomials in the field of rational functions) known to be decidable?

flag	asked Oct 6 at 15:18 Lilach Leibovich 33●4

Andreas Blass · Accepted Answer · 2012-10-06 16:48:18Z UTC

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Assuming you're working over the real numbers (or at least over a real-closed field), the answer is yes. The question lies within the first-order theory of real-closed fields, and that theory is decidable by an old theorem of Tarski.

answered Oct 6 at 16:48

Andreas Blass
32.9k●4●54●109

	hmm, yes, I forgot to mention (actually, any field which is dense in its real closure). Are there known relativley efficient algorithmm to it (i.e is it atleast primitive recursive), maybe using SOS stuff? – Lilach Leibovich Oct 6 at 17:01
	For real-closed fields, the decision procedure is primitive recursive, in fact "merely" double-exponential time (which I wouldn't call efficient). See for example en.wikipedia.org/wiki/Real_closed_field. Fields that are merely dense in their real-closures are likely to cause trouble. Even for the rationals, your question will involve arithmetical issues that I'd expect to be undecidable. – Andreas Blass Oct 6 at 23:55

Decidability of the generated order

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