Possible Duplicate:
Does the action of an affine group scheme preserve the nilradical of an algebra?

Let the group scheme $G$ act on the scheme $X$. I labored for a time under the misapprehension that the reduced subscheme $X_\text{red}$ would be automatically $G$-invariant -- it's such a canonical subscheme, how could it leave? But if $X=G$ is a nonreduced group scheme (e.g. {$g \in {\mathbb G}_m : (g-1)^p = 0$}), over a field of characteristic $p$), then this obviously fails.

What is a good condition to ensure that $G$ preserves $X_\text{red}$? Is it enough to assume $G$ is itself reduced?

EDIT: As Angelo points out in his comment, this is nearly an exact duplicate. And of a question I evidently read, since I commented on it. Sigh.


marked as duplicate by Allen Knutson, Ben Webster Aug 6 '11 at 15:30

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ I believe that the question is answered in <mathoverflow.net/questions/68366/…> (at least in the affine case, but I think that the proof can be adapted to the general case) $\endgroup$ – Angelo Aug 2 '11 at 17:01
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    $\begingroup$ Yes, if $G$ is smooth, then $G\times X_{red}$ is reduced, so the morphism $G\times X\to X$ defining the action of $G$ on $X$ induces a morphism $G\times X_{red}\to X_{red}$. $\endgroup$ – Qing Liu Aug 2 '11 at 19:17