Timeline for Is a specific endomorphism of $A_1$ an automorphism?
Current License: CC BY-SA 4.0
16 events
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Jul 4, 2019 at 16:01 | review | Suggested edits | |||
Jul 4, 2019 at 16:35 | |||||
Jul 4, 2019 at 12:57 | history | edited | user237522 | CC BY-SA 4.0 |
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Jul 4, 2019 at 12:51 | comment | added | user237522 | @YCor, thank you for your comment. ok, I will change the notation. Indeed, "$y$ does not divide $A$" means $A \notin A_1y$. | |
Jul 4, 2019 at 12:48 | comment | added | YCor | $A$ is not a good notation for an element of a ring $A_1$... does "$y$ does not divide $A$" mean $A\notin A_1y$? "divide" is somewhat ambiguous in a noncommutative ring. | |
Jul 4, 2019 at 12:45 | history | edited | YCor |
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Jul 4, 2019 at 12:09 | history | edited | user237522 | CC BY-SA 4.0 |
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Jul 4, 2019 at 12:08 | comment | added | user237522 | @JohnOmielan, thank you for your comment. You are right. I will add the link. | |
Jul 4, 2019 at 0:29 | comment | added | John Omielan | @user237522 Thanks for mentioning you asked the question also in MSE, but I suggest in the future you also include the link, e.g., as in here, to make it easier for anybody else to check on it. Thanks. | |
Jul 3, 2019 at 15:59 | comment | added | user237522 | Thank you for your advice. (I have tried considering the $(1,-1)$-degrees. Perhaps I should try more). | |
Jul 3, 2019 at 12:41 | comment | added | Marco Farinati | did you try the Z-grading $|x|=1$ and $|y|=-1$, introducing $h=yx$, and writing $A$ and $B$ in terms of polynomials on $h$ and powers of $x$ + polynomial on $h$ times powers of $y$? Bavula had a lot of success using this grading | |
Jul 3, 2019 at 9:41 | comment | added | user237522 | @MarcoFarinati, thanks. I begin with a specific $f: (x,y) \mapsto (Ay,x+By)$. Yes, I am familiar with the Dixmier Conjecture. My above question is about a special case. | |
Jul 2, 2019 at 22:03 | comment | added | Marco Farinati | Maybe I didn't understand your question properly. Do you begin with an endomorphism $f$ or you begin with $A$ and $B$ and try to define $f$? You probable know that $f$ Endo implies $f$ auto is an old conjecture | |
Jul 2, 2019 at 21:28 | comment | added | user237522 | @MarcoFarinati, thank you for your comment (but $[q,p] \neq 1$, as you mentioned). | |
Jul 2, 2019 at 21:24 | comment | added | Marco Farinati | If $B=0$ and $A=x$ then $y$ does not divide $A$, but relations are not preserved | |
Jul 2, 2019 at 20:06 | history | edited | user237522 | CC BY-SA 4.0 |
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Jul 2, 2019 at 20:00 | history | asked | user237522 | CC BY-SA 4.0 |