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Let $X$ be a variety over $k$ of characteristic $p>0$ (you can assume $k$ algebraically closed and $X$ normal) with an action of the group scheme of $p$-th roots of unity $\mu_p = {\rm Spec}\ k[\varepsilon]/(\varepsilon^p - 1)$. Denote the quotient $X/\mu_p$ by $Y$ and the quotient morphism $X\to Y$ by $\pi$.

Let $D$ be a Weil divisor on $X$. Is the sheaf $\pi_* (\mathscr{O}_X(D))$ a direct sum of sheaves of the form $\mathscr{O}_Y(E)$ for some Weil divisors $E$ on $X$?

For an action of $\mu_n$, $n$ not divisible by $p$, we get the above statement because the push-forward decomposes according to the characters of $\mu_n$, and if we assume that there is an orbit on which $\mu_n$ acts freely, then all these summands are nonzero - hence all have rank one and are reflexive. So we need a form of ,,diagonalizability'' for $\mu_p$. I don't see how the proof for $\mu_n$ above could be translated to a proof for $\mu_p$.

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  • $\begingroup$ (1) What do you mean by $\mathcal{O}_X(D)$ if $D$ is not a Cartier divisor? (2) Do you assume $D$ invariant under $\mu_p$? $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 3, 2010 at 12:16
  • $\begingroup$ Dear Laurent, sorry for the delay. (1) $O_X(D)$ is the reflexive sheaf, the subsheaf $D+(f)\geq 0$ of the sheaf of rational functions, (2) Yes, you are right, I should assume $D$ stable under the action. $\endgroup$
    – Piotr Achinger
    Commented Dec 11, 2010 at 20:15

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If you cannot generalize, you may try to simplify:-)) Representations of $\mu_p$ are still completely reducible: they are just graded vector spaces by the cyclic group of order $p$. The first part of your argument goes through.

For the second part, you have a simplification as $\mu_p$ has no subgroup schemes! So your stabilisers are either trivial of full $\mu_p$. If the action is nontrivial, there is a point with trivial stabiliser, and there you know that each simple character of $\mu_p$ appears.

Did I miss anything?

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  • $\begingroup$ Dear Bugs, thanks for your answer! Your argument sounds ok. As I understand, we replace the eigensheaves $g\cdot s = \chi(g)\cdot s$ by the subsheaves consisting of all sections $s$ such that $\mu^* (s) = s\otimes x^i$ (for $i=0, \ldots, p-1$), where $\mu : X\times \mu_p \to X$ is the action and $\mu_p = \mathrm{Spec}\, k[x]/(x-1)^p$. Is this correct? $\endgroup$
    – Piotr Achinger
    Commented Dec 11, 2010 at 20:20
  • $\begingroup$ Yes, this is correct. $\endgroup$
    – Bugs Bunny
    Commented Dec 12, 2010 at 16:00

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