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Dec 23, 2022 at 11:08 history edited YCor CC BY-SA 4.0
formatting
Dec 23, 2022 at 10:42 comment added David E Speyer @Vincent Sorry for introducing more confusion. I did, indeed, want $n$ nilpotent.
Dec 23, 2022 at 10:41 history edited David E Speyer CC BY-SA 4.0
edited body
Dec 23, 2022 at 9:14 comment added Vincent It seems to me that the original formulation was the right one and the edit introduced an error. But then again I know virtually nothing about this subject, so it seems likely that I am wrong. Either way I would be happy if someone could clarify this
Dec 23, 2022 at 9:07 comment added Vincent I am confused now about the $x$ nilpotent vs $n$ nilpotent business. The way it is written now in the post it looks like the homomorphism is very picky and only accepts nilpotent elements as input. But that is not how homormorphisms are supposed to work, right? The domain should be the whole group
Dec 23, 2022 at 4:16 comment added LSpice @YCor, ah, you are right. I thought the $n$ in $f_i = n^i/i!$ was an integer; it didn't occur to me that it was a ring element.
Dec 23, 2022 at 3:13 comment added YCor @LSpice No, I think "$n$ is nilpotent" is correct. You have a unital ring homomorphism $R[y,y^{-1}]\to R[x]$, mapping $y$ to an element $\sum_{i\ge 0}f_ix^i$; this element must be inversible and this is equivalent to: $f_0$ is inversible in $R$ and $f_i$ is nilpotent for all $i\ge 1$. (Then the condition of being a Hopf algebra homomorphism implies the given formula, namely $f_i=f_1^i/i!$, with $f_1$ nilpotent.)
Dec 23, 2022 at 1:35 history edited David E Speyer CC BY-SA 4.0
deleted 5 characters in body
Dec 22, 2022 at 23:03 comment added LSpice "n must by nilpotent" → "$x$ must be nilpotent", I think.
S Jan 11, 2021 at 11:54 history suggested Captain Lama CC BY-SA 4.0
Put all the Hom, Spec, and G in TeX format.
Jan 11, 2021 at 10:57 review Suggested edits
S Jan 11, 2021 at 11:54
S Apr 30, 2019 at 16:55 history suggested display llvll CC BY-SA 4.0
just added some dollars
Apr 30, 2019 at 16:46 review Suggested edits
S Apr 30, 2019 at 16:55
Nov 12, 2017 at 22:12 comment added Qiaochu Yuan "Characteristic zero" here should be interpreted to mean "$\mathbb{Q}$-algebra," I guess.
Nov 4, 2009 at 17:30 vote accept David Zureick-Brown
Nov 4, 2009 at 15:58 history answered David E Speyer CC BY-SA 2.5