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Suppose $X/k$ is a finite commutative group scheme over a perfect field. Then we know that the category $\mathcal{N}$ of finite commutative group schemes over $k$ is abelian and isomorphic to a direct product $$\mathcal{N} = \mathcal{N}_{loc} \times \mathcal{N}_{red} = \mathcal{N}_{ll} \times \mathcal{N}_{lr} \times \mathcal{N}_{rr} \times \mathcal{N}_{rl}$$ where $\mathcal{N}_{loc}$, $\mathcal{N}_{red}$ denote the full subcategories of $\mathcal{N}$ corresponding to objects whose corresponding rings are either reduced or local respectively. One can describe this at the level of bialgebras, apparently it corresponds to a "direct sum" of bialgebras.

$\textbf{Question:}$ Given a finite commutative group scheme $G$ one can of course calculate the reduced and local components, but how do you determine the local-local, local-reduced etc., components? It's not clear to me if this is easy from formal properties of duality. Is it true that $G \cong G_{ll} \times_k G_{rr} \times_k G_{lr} \times_k G_{rr}$.

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  • $\begingroup$ What's local-local and local-reduced? $\endgroup$
    – Will Sawin
    Sep 23, 2013 at 3:33
  • $\begingroup$ @Will: local-local means $G$ and its Cartier dual are connected (equivalently local, by $k$-finiteness), and local-reduced means $G$ is connected and its Cartier dual is etale (as $k$ is perfect). $\endgroup$
    – Marguax
    Sep 23, 2013 at 4:42
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    $\begingroup$ @none: The connected-etale sequence of the commutative $G$ uniquely splits by Galois descent from the same over $\overline{k}$ (as $k$ is perfect), and likewise for the Cartier dual, so you thereby get the resulting canonical (even unique) 4-fold direct product decomposition of $G$. That is the entire content underlying the proof of your second sentence, and is all explained in Mumford's book on abelian varieties (over $\overline{k}$, so then use Galois descent via uniqueness). How do you understand your 2nd sentence if you don't know the answer to the final sentence in your question? $\endgroup$
    – Marguax
    Sep 23, 2013 at 4:44
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    $\begingroup$ @Marguax,you're right it is the entire content of the proof of the decomposition, I guess that is what I'm trying to understand, but I think I get it now. Suppose $G$ is reduced, then we can of course calculate a decomposition of the dual $G^D = G_{rl} \times_k G_{rr}$. Dualizing takes tensor products to tensor products so we have our decomposition $G = G_{rr} \times G_{rl}$. I was confused about how dual behaves with tensor products, but that follows from the definitions. $\endgroup$
    – none
    Sep 23, 2013 at 13:09

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