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Jan 25, 2021 at 18:11 answer added Arno Fehm timeline score: 4
Jun 4, 2019 at 2:13 answer added Thomas Scanlon timeline score: 15
Jun 2, 2019 at 8:34 comment added YCor @JoelDavidHamkins because when there's such an immediate answer, I expect from the OP a reaction such as "oh, I meant a finite algebraic extension".
Jun 2, 2019 at 8:27 comment added Joel David Hamkins @YCor Why not post your comment as an answer? You have answered question 1, the point being that Tarski's elimination of quantifiers argument applies in any real-closed field, including the real algebraic numbers. So that is an algebraic extension in which $\mathbb{Z}$ is not definable, because the theory is decidable.
Jun 1, 2019 at 15:24 comment added YCor Without further assumption on the algebraic extension, $\mathbf{Z}$ is not definable (even with constants) in the real-closed field $\mathbf{R}\cap\bar{\mathbf{Q}}$.
Jun 1, 2019 at 14:33 history edited YCor CC BY-SA 4.0
clarified question, added tag
Jun 1, 2019 at 14:24 comment added George Peterzil @YCor Yes, that is what I mean
Jun 1, 2019 at 14:21 comment added YCor Also "undecidable", does it mean here that the there's no Turing machine whose input is a free formula in the language of fields and whose output is yes/no according to whether it's true in $\mathbf{Q}$?
Jun 1, 2019 at 14:15 comment added YCor The 1st § sounds a bit vague. Do you mean Robinson proved that there is a formula $F(x)$ in the language of fields such that for $x\in\mathbf{Q}$ we have $F(x)$ $\Leftrightarrow$ $x\in\mathbf{Z}$? In the question 1, by "the integers" do you mean $\mathbf{Z}$ or the ring of integers in $K$?
Jun 1, 2019 at 14:10 review First posts
Jun 1, 2019 at 15:12
Jun 1, 2019 at 14:09 history asked George Peterzil CC BY-SA 4.0