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How can I prove that the Cartier dual of αp is again αp (using the Yoneda lemma)? It should be something like $\alpha_p(R) \to (\alpha_p(R) \to \mu_p(R)),x \mapsto (y \mapsto exp_{p−1}(x+y)$, where $exp_{p−1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.

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The formula is $x\mapsto (y\mapsto\exp_{p-1}(xy))$. – Torsten Ekedahl Nov 11 '10 at 11:48
Thanks! How do I see this is an isomorphism? – Timo Keller Nov 11 '10 at 16:36
Why is it important to prove this using the Yoneda lemma? – stankewicz Nov 11 '10 at 16:47
up vote 6 down vote accepted

It is probably a bad idea to try to compute the Cartier dual but better to let Cartier do that for you... If $G$ is a flat commutative finite group scheme with affine algebra, the commutative and cocommutative $A$ which is the flat over the base ring $R$. Then the Cartier dual is the spectrum of the dual Hopf algebra $A^\ast$ of $A$. The proof of this is simple enough; an $R$-algebra homomorphism $A\rightarrow R$ corresponds to a $\varphi\in A^\ast$ of multiplicative type, $\Delta^\ast(\varphi)=\varphi\otimes\varphi$, which in turn corresponds to a Hopf algebra map $R[t,t^{-1}]\rightarrow A^\ast$. As this can be done for all $R$-algebras we get an isomorphism of functors.

Doing this for $\alpha_p$ which has $A=R[x]/(x^p)$ we get that $A^\ast$ has a basis dual to $x^i$ of the form $1/i!\partial^i/\partial x^i$. Unravelling the definitions one gets the formula $s\mapsto(t\mapsto \exp_{p-1}(st))$.

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very nice! cool! – SGP Apr 16 '11 at 11:39

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