Global centralizers in Jordan-Chevalley decomposition in bad characteristic

Let $G$ be an affine algebraic group defined over an algebraically closed field $k$ of arbitrary characteristic, and write $\mathfrak{g}$ for its Lie algebra. Given $X\in\mathfrak{g}$, it has (relative to $G$) its Jordan-Chevalley decomposition $X=s+n$. Since the decomposition is preserved by morphisms of affine groups, conjugation by any $k$-point of the centralizer $C_G(X)$ will preserve the decomposition, so $C_G(X)(k)$ is contained in $C_G(s)(k)$.

If we consider the centralizers as group schemes over $k$, then is $C_G(X)$ contained in $C_G(s)$? I'm mostly interested in the case of $G$ reductive. Useful references would also be appreciated.

The answer is affirmative for $G$ any smooth affine $k$-group. We may assume $p = {\rm{char}}(k) > 0$. By 14.26 in Borel's textbook on algebraic groups, $n$ is in the Lie algebra of a unipotent subgroup $U$ of $G$. The $p$-operator $Y \mapsto Y^{[p]}$ on the Lie algebra of a smooth affine group (or more generally group scheme of finite type) over $k$ is functorial in the group, and it vanishes for the Lie algebra of $\mathbf{G}_{\rm{a}}$, so via a suitable composition series on $U$ we see that $Y \mapsto Y^{[p^m]}$ kills ${\rm{Lie}}(U)$ for large $m$, and hence kills the nilpotent $n$. But since $[s,n]=0$, we have $X^{[p^i]} = s^{[p^i]} + n^{[p^i]}$ for all $i$, so $X^{[p^m]} = s^{[p^m]}$.
The theory of the $p$-operation works on Lie algebras of arbitrary group schemes over arbitrary $\mathbf{F}_p$-algebras (it is defined as the $p$-power on left-invariant derivations on the structure sheaf linear over the base ring), so by reasoning with Yoneda's Lemma we see that as closed subgroup schemes of $G$ we have $$C_G(X) \subset C_G(X^{[p^m]}) = C_G(s^{[p^m]}).$$ Thus, to conclude that $C_G(X) \subset C_G(s)$ it suffices to show that $C_G(s) = C_G(s^{[p]})$ for any semisimple $s \in \mathfrak{g}$. Using an inclusion $G \hookrightarrow {\rm{GL}}_N$ reduces this to the case $G = {\rm{GL}}_N$ and diagonal $s \in \mathfrak{gl}_N = {\rm{Mat}}_N(k)$. Consideration of behavior with respect to weight-space decomposition then does the job since distinct elements of $k$ have distinct $p$th powers.
• I think that your argument about unipotent subgroups (of arbitrary $G$) is overkill; as with semisimple elements, you can show that $n^{[p]^N} = 0$ for $N$ sufficiently large simply by embedding $G$ in $\mathrm{GL}_N$. – LSpice Jan 14 '16 at 18:17