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Mar 11, 2020 at 17:37 vote accept user237522
Mar 11, 2020 at 0:51 history edited LSpice CC BY-SA 4.0
Minor proofreading
Mar 11, 2020 at 0:13 answer added Johannes Hahn timeline score: 7
Mar 10, 2020 at 20:34 comment added user237522 Thank you for your comment. (You can write it as an answer if you like).
Mar 10, 2020 at 12:37 comment added Johannes Hahn Your argument for $n=2$ generalises and shows that $x_1$ is always mapped to $\lambda x_1^k$. Therefore if you consider the subring $S:=k[x_1^{\pm 1}]$, a general $k$-linear automorphism of is a composition of a $S$-linear automorphism of $S[x_2,...x_n]$ and an automorphism of $S$ (acting on $S[x_2,...,x_n]$ by acting on coefficients).
Mar 10, 2020 at 7:52 history edited YCor
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May 12, 2017 at 11:41 history migrated from math.stackexchange.com (revisions)
May 7, 2017 at 11:07 history asked user237522 CC BY-SA 3.0