Timeline for Generalizations of Abhyankar-Moh theorem (embeddings of the line in the plane)
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2018 at 11:39 | comment | added | Matthias Wendt | For the base field in the Abhyankar-Moh(-Suzuki) theorem, the following statements are mentioned at the beginning of Section 4 of van den Essen's survey (mentioned in the answer): any field of characteristic zero works, one can even take integral domains containing $\mathbb{Q}$, but there are counterexamples in positive characteristic. | |
| Apr 10, 2018 at 6:27 | comment | added | user237522 | Please, for Abhyankar-Moh(-Suzuki) theorem $k=1$, $n=2$, what is the base field? | |
| Mar 27, 2018 at 18:26 | comment | added | user237522 | ok, thanks (interesting what Srinivas notes). | |
| Mar 27, 2018 at 18:22 | comment | added | Matthias Wendt | @user237522: sure, I should have said arbitrary infinite field (but Srinivas notes that he has no example where his result fails over finite fields). | |
| Mar 26, 2018 at 22:03 | comment | added | user237522 | Thank you very much! (I guess that you have meant in your last comment "for Srinivas' result the base field can be an arbitrary infinite field etc.", as you have mentioned in your answer). | |
| Mar 26, 2018 at 21:34 | comment | added | Matthias Wendt | @user237522: for Srinivas' result, the base field can be an arbitrary field, not necessarily closed, not necessarily characteristic zero. I am not aware of generalizations to rings. It would seem to me that this is a much more complicated question. | |
| Mar 26, 2018 at 10:23 | comment | added | user237522 | Thank you for answering me. So the base field can be an arbitrary field? (1) not necessarily algebraically closed? (2) not necessarily of characteristic zero? (3) an integral domain? (4) an arbitrary commutative ring? Is it true that the answers are: (1) yes. (2) yes. (3) no? or maybe considering its field of fractions may yield something interesting?. (4) no? | |
| Mar 26, 2018 at 9:08 | comment | added | Matthias Wendt | $\mathbb{A}^n$ denotes the affine space of dimension $n$ (base field here is implicit). | |
| Mar 25, 2018 at 5:02 | comment | added | user237522 | Please, what is $\mathbb{A}$? Is it ok to take $\mathbb{A}=\mathbb{R}$? | |
| Mar 19, 2018 at 14:45 | vote | accept | user237522 | ||
| Mar 19, 2018 at 14:01 | history | answered | Matthias Wendt | CC BY-SA 3.0 |