Does the action of an affine group scheme preserve the nilradical of an algebra? 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.
 A: 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 $(x) \otimes \bigl(k[x]/(x^p)\bigr)$.
A: 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.
