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Let $G = \operatorname{Spec} R$ be a finite group scheme, with $R$ a finite-dimensional Hopf algebra. By Tannakian duality, we should be able to reconstruct $G$ from the category of $G$-representations together with its forgetful functor and tensor product.

It is known that the category of $G$-representations is equivalent to the category of $R$-comodules, which is also equivalent to the category of $R^*$-modules (see Jantzen, Representations of Algebraic Groups, 8.6).

Now, consider two finite group schemes $\alpha_p$ and $\mathbb{Z}/p\mathbb{Z}$. Although these are not isomorphic, their dual coordinate rings $k[\alpha_p]^*$ and $k[\mathbb{Z}/p\mathbb{Z}]^*$ are both isomorphic to $k[t]/(t^p)$. This implies that the categories of $\alpha_p$-representations and $\mathbb{Z}/p\mathbb{Z}$-representations are both equivalent to the category of $k[t]/(t^p)$-modules.

To avoid contradicting Tannakian duality, the categories of $\alpha_p$- and $\mathbb{Z}/p\mathbb{Z}$-representations should have different forgetful functor or tensor product structures. However, the category of $k[t]/(t^p)$-modules also has a natural forgetful functor and tensor product that appear to match those of the original categories.

This seems contradictory. I would appreciate any insights into where the issue might be.

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    $\begingroup$ The tensor structure is not correct: on $Mod_{R^*}$, it crucially depends on the coalgebra structure on $R^*$, i.e. the algebra structure on $R$ $\endgroup$ Commented Nov 4 at 22:26
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    $\begingroup$ While both categories are equivalent to $k[t]/(t^p)$-modules, this equivalence is not monoidal. Indeed, the monoidal structure could not be $- \otimes_{k[t]/t^p} -$, because then the forgetful functor to $k$-vector spaces is not monoidal (which is a key condition to run the Tannakian machine). $\endgroup$ Commented Nov 4 at 22:27
  • $\begingroup$ @R.vanDobbendeBruyn I suggest to upgrade your comment to an answer. $\endgroup$ Commented Nov 4 at 23:10
  • $\begingroup$ @MaximeRamzi I suggest to upgrade your comment to an answer. $\endgroup$ Commented Nov 4 at 23:11

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$\newcommand{\Z}{\mathbf{Z}}\DeclareMathOperator{\spec}{Spec}\DeclareMathOperator{\Rep}{Rep} \DeclareMathOperator{\QCoh}{QCoh}$The Cartier dual of $\alpha_p$ is itself, while the Cartier dual to $\Z/p$ is $\mu_p$. While $\alpha_p$ and $\mu_p$ are isomorphic as schemes, they are not isomorphic as group schemes. The category $\Rep(\alpha_p)$ (resp. $\Rep(\Z/p)$) with its standard symmetric monoidal structure is therefore equivalent to $\QCoh(\alpha_p)$ (resp. $\QCoh(\mu_p)$) with the convolution tensor structure (and this depends on the group structures on $\alpha_p$ and $\mu_p$). This fixes the "contradiction": you need to use the symmetric monoidal structure on your category to run Tannakian reconstruction.

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