Timeline for The $1$-dimensional Jacobian Conjecture over $\mathbb{Z}$-torsion free rings
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| May 20, 2023 at 12:08 | vote | accept | M.G. | ||
| May 20, 2023 at 6:55 | history | edited | M.G. | CC BY-SA 4.0 |
strong characteristic 0 replaced by a more standard name
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| May 20, 2023 at 6:43 | comment | added | M.G. | @R.vanDobbendeBruyn: You are right, $\mathbb{Z}$-torsion free is much better than my made-up non-standard name! I will edit accordingly. | |
| May 20, 2023 at 0:27 | answer | added | R. van Dobben de Bruyn | timeline score: 5 | |
| May 19, 2023 at 22:30 | comment | added | R. van Dobben de Bruyn | Probably the most common name for your current condition is that $R$ is $\mathbf Z$-torsion free (equivalently, flat over $\mathbf Z$, whereas characteristic $0$ only means that $\operatorname{Spec} R \to \operatorname{Spec} \mathbf Z$ is dominant). | |
| May 19, 2023 at 16:20 | history | edited | M.G. | CC BY-SA 4.0 |
added some characteristic clarifications
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| May 19, 2023 at 15:35 | history | edited | M.G. | CC BY-SA 4.0 |
added 30 characters in body
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| May 19, 2023 at 15:23 | history | edited | M.G. | CC BY-SA 4.0 |
added 16 characters in body
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| May 19, 2023 at 15:16 | history | edited | M.G. | CC BY-SA 4.0 |
deleted 41 characters in body
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| May 19, 2023 at 15:16 | comment | added | M.G. | I've modified the question b/c "equicharacteristic" does not seem suitable since $\mathbb{Z}$ is of mixed characteristic and is allowed. The correct characteristic assumption should be that $R$ admits an extension containing $\mathbb{Q}$ in order to avoid pathological examples like Z x F_p. | |
| May 19, 2023 at 15:11 | history | edited | M.G. | CC BY-SA 4.0 |
deleted 26 characters in body; edited title
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| May 19, 2023 at 14:53 | history | edited | M.G. | CC BY-SA 4.0 |
added 84 characters in body
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| May 19, 2023 at 14:40 | history | edited | M.G. | CC BY-SA 4.0 |
characteristic -> equichar.
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| May 19, 2023 at 14:28 | comment | added | R. van Dobben de Bruyn | But for $R = \mathbf Z \times \mathbf F_p$ it should be false, right? Take $f = (0,1)x^p-(1,1)x$. | |
| May 19, 2023 at 14:24 | comment | added | R. van Dobben de Bruyn | Does characteristic $0$ for you mean equicharacteristic $0$ (i.e. $R$ contains $\mathbf Q$), or is $R = \mathbf Z$ allowed as well? What about $R = \mathbf Z \times \mathbf F_p$? | |
| May 19, 2023 at 14:10 | history | edited | M.G. | CC BY-SA 4.0 |
added 16 characters in body
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| May 19, 2023 at 14:05 | history | asked | M.G. | CC BY-SA 4.0 |