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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