Timeline for Could the Jacobian conjecture be undecidable?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 25, 2013 at 1:55 | vote | accept | Marty | ||
| May 17, 2013 at 17:34 | answer | added | Timothy Chow | timeline score: 5 | |
| May 17, 2013 at 14:25 | comment | added | Emil Jeřábek | Yes, if it is undecidable in a half-decent theory, then it is true. Yes, you cannot prove in, say, ZFC that it is undecidable in ZFC, but then again, you cannot prove in ZFC that anything is undecidable in ZFC. However, it is conceivable that the undecidability of the conjecture in ZFC is provable by assuming some stronger hypothesis, such as the consistency of ZFC. | |
| May 16, 2013 at 3:56 | comment | added | Steven Landsburg | Felipe: If PA is consistent, then the Godel sentence is automatically true for trivial reasons, though it can't be proved in PA. | |
| May 16, 2013 at 3:07 | comment | added | Felipe Voloch | @Steven: My logic is rusty but I don't think the Godel sentence is automatically true. But I may be wrong. Of course "undecidable" is within a set of axioms and something undecidable in PA may be provable in ZFC. Hopefully, one of the many experts will chime in. | |
| May 16, 2013 at 1:30 | comment | added | Steven Landsburg | Felipe Voloch: It's clear from my earlier comment that I'm not at my best tonight, so maybe I'm missing something fundamental again, but: The Godel sentence for Peano Arithmetic is undecidable and therefore true, but it doesn't follow that I can't prove it's undecideable. | |
| May 16, 2013 at 1:15 | comment | added | Steven Landsburg | Henry Cohn: Ah. Right. I forgot that the first order theory of fields can't talk about natural numbers. I knew I was missing something! | |
| May 16, 2013 at 1:08 | comment | added | Felipe Voloch | If it is undecidable, then it is true, since if it were false, your program would halt for some n with a proof that it is false. So you can't prove that it is undecidable. | |
| May 16, 2013 at 0:57 | comment | added | Henry Cohn | @Steven Landsburg: For each fixed $n$ it is, but you can't quantify over $n$ in the first-order language of fields. | |
| May 16, 2013 at 0:45 | comment | added | Steven Landsburg | I'm probably missing something fundamental here, but....isn't "For all $n$, $J(3,n)$" a sentence in the theory of algebraically closed fields of characteristic zero? So doesn't completeness still apply? | |
| May 16, 2013 at 0:12 | history | asked | Marty | CC BY-SA 3.0 |