Timeline for Again, polynomial bijection $f:{\mathbb Q}\times{\mathbb Q}\times{\mathbb Q}\to {\mathbb Q}$
Current License: CC BY-SA 4.0
17 events
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| Mar 2, 2021 at 12:02 | comment | added | Giulio Bresciani | Actually, if one uses more generally the methods of my paper rather than the result as stated, the geometric Lang conjecture can be avoided. | |
| Mar 1, 2021 at 17:15 | comment | added | Giulio Bresciani | @MattF. If you assume the geometric Lang conjecture in addition to the weak Bombieri-Lang conjecture, then my result rules out the case in which the generic fiber is of general type. It remains then to address the cases in which the fiber has lower Kodaira dimension, but that's probably easy. | |
| Feb 20, 2021 at 14:46 | comment | added | user44143 | Can anyone extend Bresciani's conditional results on $\mathbb{Q}\times \mathbb{Q}\to \mathbb{Q}$ to answer this question? arxiv.org/abs/2101.01090 | |
| Apr 22, 2020 at 21:36 | comment | added | R. van Dobben de Bruyn | Maybe you can localise the category of polynomial maps at the maps that induce bijections... | |
| Apr 22, 2020 at 19:54 | comment | added | Wojowu | @R.vanDobbendeBruyn I doubt it (and I didn't mean to imply that). If there is, it definitely cannot have the obvious concretization, because the inverse of $\mathbb R\to\mathbb R,x\mapsto x^3$ is not a polynomial bijection. | |
| Apr 22, 2020 at 19:06 | comment | added | R. van Dobben de Bruyn | @Wojowu: Is there a natural category whose isomorphisms are the polynomial bijections? | |
| Apr 22, 2020 at 15:50 | history | edited | user147204 | CC BY-SA 4.0 |
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| Apr 22, 2020 at 15:34 | comment | added | Wojowu | I doubt there is a simple "formal" argument for this. In various categories (e.g. groups) it is possible to have an object $A$ isomorphic to $A^3$ but not isomorphic to $A^2$. | |
| Apr 22, 2020 at 14:16 | comment | added | Yaakov Baruch | @MichaelRozenberg. What is the presumption? Maybe OP has no fruitful attempts or promising ideas to show - that wouldn't invalidate the question. A trivial answer would, but is there one? | |
| Apr 22, 2020 at 13:50 | comment | added | Yaakov Baruch | Trivial, but worth noting that the converse holds: $\mathbb{Q}^2\leftrightarrow \mathbb{Q} \implies \mathbb{Q}^2\times\mathbb{Q}\leftrightarrow \mathbb{Q}\times\mathbb{Q}\leftrightarrow\mathbb{Q}$. | |
| Apr 22, 2020 at 12:12 | comment | added | Gro-Tsen | @Wowoju mathoverflow.net/q/21003/17064 (OP should have linked this) | |
| Apr 22, 2020 at 11:33 | comment | added | Wojowu | Why "Again" in the title? Was this question already asked before? | |
| Apr 22, 2020 at 8:11 | history | edited | Tomasz Kania | CC BY-SA 4.0 |
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| Apr 22, 2020 at 7:27 | history | edited | user147204 | CC BY-SA 4.0 |
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| Apr 22, 2020 at 4:38 | comment | added | Michael Rozenberg | What do you think about it? Show please your attempts. | |
| Apr 22, 2020 at 4:36 | review | First posts | |||
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| Apr 22, 2020 at 4:34 | history | asked | user147204 | CC BY-SA 4.0 |