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Feb 11, 2016 at 15:04 vote accept Thiago
Jan 29, 2015 at 21:17 comment added ACL In case 3), I believe that $t\mapsto 2-t$ induces an automorphism of $I_f$.
Jan 29, 2015 at 17:08 comment added Johannes Hahn No, I really mean $\alpha(f)^{-1} = \frac{1}{\alpha(f)}$.
Jan 29, 2015 at 17:06 answer added Eric Wofsey timeline score: 11
Jan 29, 2015 at 16:56 comment added Thiago 0. For us an algebra does not need to be unitary. 1. It is just for that particular example. In the general case, they are generated by $\{t^if | 0\leq i< \deg(f)\}$, as you mentioned. 2. You mean $\alpha^{-1}(f)$? I did not understand.
Jan 29, 2015 at 16:50 history edited Daniel Loughran
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Jan 29, 2015 at 16:45 comment added Johannes Hahn 0. An "algebra" has an unit in my book, but whatever... 1. Are you sure that $f$ and $tf$ generate $I_f$ ? Isn't $\{t^i f \mid 0\leq i<deg(f)\}$ a minimal generating system (consider$f$-adic expansions)? Or was this just for that particular example? 2. It seems to me that $Aut(I_f)$ embeds into $Aut(\mathbb{C}[t])$ via $\alpha \mapsto (t \mapsto \alpha(f)^{-1}\alpha(tf))$. This could be useful.
Jan 29, 2015 at 16:15 history edited Thiago
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Jan 29, 2015 at 16:07 history asked Thiago CC BY-SA 3.0