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Let $k$ be a commutative ring and let $G$ be a flat affine algebraic group scheme over $k$. Let $G$ act by algebra automorphisms on the commutative $k$-algebra $A$. So $G(R)$ acts by $R$-algebra automorphisms on $A\otimes_k R$ for any commutative $k$-algebra $R$. Let $N$ be the nilradical of $A$. Is $N$ always a $G$ submodule? So is the image of $N\otimes_k R$ in $A\otimes_k R$ invariant under $G(R)$? I only need it when $G$ is a Chevalley group scheme, in which case it is true. But my question is if this is a general fact.

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2 Answers 2

up vote 13 down vote accepted

This is true if you assume that $G$ is smooth. Consider the coaction $A \to A \otimes_k k[G]$; since $k[G]$ is a smooth $k$-algebra, the nilradical of $A \otimes_k k[G]$ is $N \otimes_k k[G]$; since $N$ is sent to the nilradical of $A \otimes_k k[G]$, this implies the thesis.

Otherwise it is false in general, even over algebraically closed fields. For example, take $G = \alpha_p = \mathop{\rm Spec}k[x]/(x^p)$ in characteristic $p$, and consider the action of $G$ over itself by translation. This corresponds to the coaction $k[x]/(x^p) \to \bigl(k[x]/(x^p)\bigr) \otimes_k \bigl(k[x]/(x^p)\bigr)$ sending $x$ into $x\otimes 1 + 1 \otimes x$. The nilradical of $k[x]/(x^p)$ is $(x)$; but $x\otimes 1 + 1 \otimes x$ is not in $(x) \otimes \bigl(k[x]/(x^p)\bigr)$.

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Thank you much! –  Wilberd van der Kallen Jun 21 '11 at 13:14

I think this is probably not true even for $k$ a field. There exist finite commutative group schemes $G$ (over imperfect fields) such that $G_{red}$ is not a subgroup scheme of $G$. If we let $A$ be the coordinate ring of $G$ (with the regular action) then I think this might imply that the nilradical of $A$ will not be preserved by the $G$ action.

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Can one exhibit a simple example of such a beast? –  Mariano Suárez-Alvarez Jun 21 '11 at 13:00
    
I'm pretty sure that BCnrd did so in a comment somewhere, but he may have only mentioned its existence (and I'm not finding either from a cursory search). –  Allen Knutson Jun 21 '11 at 13:43
    
@Mariano Here is a natural, though not so simple, example: Let $k$ be any field of char $p>0$ and $k(t)$ the field of rational functions over $k$. Let $E$ be any elliptic curve over $k(t)$ with $j$-invariant $t$. Then $E[p]$, the $p$-torsion subscheme of $E$, has the desired property. This can be deduced from the discussion in mathoverflow.net/questions/17101 which is perhaps also what Allen Knutson is referring to. (Given Angelo's answer this is somewhat moot so I don't give any more details.) –  ulrich Jun 21 '11 at 14:33
    
Just to spell out: Angelo says that in the $G$ action on $G$, $G$ might not preserve $G_{red}$. ulrich says that even $G_{red}$ might not preserve $G_{red}$, which is obviously worse but not actually necessary for answering Wilberd's question. –  Allen Knutson Aug 2 '11 at 17:49

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