Timeline for Model-completeness of real exponential fields
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2022 at 13:49 | history | edited | user44143 | CC BY-SA 4.0 |
minor corrections at beginning
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| Aug 29, 2022 at 13:16 | comment | added | user44143 | I redid this with more generality and a clearer assumption of minimality. So if there are more equations than unknowns, then some of the equations are not needed. Similarly, this assumption of minimality avoids equations of the form $f^2+g^2=0$, in favor of a pair of equations of the form $f=0, g=0$; and it avoids equations of the form $fg=0$ in favor of using $f=0$ by itself or using $g=0$ by itself. I hope that in this way, if algebraically dependent equations occur at some point, it will be possible to use some analysis or some o-minimality.to prove that they have multiple solutions. | |
| Aug 29, 2022 at 13:12 | history | edited | user44143 | CC BY-SA 4.0 |
replaced and re-placed assumption of minimality
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| Aug 29, 2022 at 13:02 | comment | added | Emil Jeřábek | Why do you assume you have $n$ equations in $n$ unknowns? You may well have $n$ equations in $m>n$ unknowns (or vice versa). | |
| Aug 29, 2022 at 13:00 | history | edited | user44143 | CC BY-SA 4.0 |
corrected proof of lemma
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| Aug 29, 2022 at 12:50 | history | edited | user44143 | CC BY-SA 4.0 |
gave algorithm in more generality
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| Aug 27, 2022 at 13:22 | history | edited | user44143 | CC BY-SA 4.0 |
added examples in response to comments
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| Aug 27, 2022 at 13:17 | history | edited | user44143 | CC BY-SA 4.0 |
added examples in response to comments
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| Aug 27, 2022 at 8:28 | comment | added | Emil Jeřábek | The example is not entirely representative as it does not involve any variables besides $u$. I included in the question another normal form for existential definitions that may be more useful for this approach. But anyway, when I tried to think about how to generalize the argument, I believe the major problem is how to show that the algebraic equations are independent; or rather, how to deal with the fact that they are not independent in general. This is unavoidable, as after all $e$ does satisfy plenty of existential formulas (but, presumably, not as a unique solution). | |
| Aug 26, 2022 at 18:45 | comment | added | user44143 | Yes, I’ve corrected this. I think we can prove this properly for arbitrary terms by starting with a large list of variables and equations, and repeatedly eliminating complicated terms via Schanuel’s conjecture. And before that, perhaps someone can either simplify the argument here, or provide a reference for iterating Schanuel’s conjecture to cite for the details. | |
| Aug 26, 2022 at 17:59 | history | edited | user44143 | CC BY-SA 4.0 |
corrected
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| Aug 26, 2022 at 17:31 | comment | added | Emil Jeřábek | Thank you. You mean the third condition is false, right? The suggestion that it contradicts Schanuel's conjecture sounds plausible, but I'm worried that proving this properly for arbitrary terms rather than for a particular example might be difficult. | |
| Aug 26, 2022 at 16:38 | history | edited | user44143 | CC BY-SA 4.0 |
removed some repetitive text and clarified the version of Schanuel’s conjecture being used
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| Aug 26, 2022 at 16:05 | history | edited | user44143 | CC BY-SA 4.0 |
made the logic more explicit and restated one algebraic relationship
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| Aug 26, 2022 at 14:01 | history | answered | user44143 | CC BY-SA 4.0 |