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Mar 23, 2021 at 18:34 answer added Neil Strickland timeline score: 2
Oct 22, 2020 at 18:08 vote accept kiran
Sep 23, 2020 at 2:56 answer added Lubin timeline score: 2
Sep 22, 2020 at 22:54 comment added kiran @Lubin So $u\cdot f$ "is" $T$ - it is not quite an endomorphism of $F$ but rather sends $F$ to $\phi^*F$. And instead of $G/\text{ker}T\cong G$ all I have is $G/\text{ker} T\hookrightarrow \phi^*G$.
Sep 22, 2020 at 17:47 comment added Lubin Your description of specialness in the Edit seems to be demanding that there would be an endomorphism $u\cdot f$ of $F$, vanishing on the kernel of $T$, so that in some sense $G/(\ker T)\cong G$ as formal groups. Have I misconstrued?
Sep 22, 2020 at 7:54 history edited YCor CC BY-SA 4.0
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Sep 22, 2020 at 7:48 history edited kiran CC BY-SA 4.0
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Sep 22, 2020 at 7:26 comment added kiran @QiaochuYuan It's a closed formal subscheme that's finite and free and is also a group. That's equivalent to the data of a monic polynomial $f(x)$ with nilpotent lower order coefficients such that $f(F(x,y))=0$ mod $(f(x),f(y))$.
Sep 22, 2020 at 5:35 comment added Qiaochu Yuan What's a finite subgroup of a formal group?
Sep 22, 2020 at 3:21 review First posts
Sep 22, 2020 at 5:11
Sep 22, 2020 at 3:18 history asked kiran CC BY-SA 4.0