Let $G$ be a reductive linear algebraic group defined over an algebraically closed field $k$ of arbitrary characteristic, and write $\mathfrak{g}$ for its Lie algebra. The Jordan-Chevalley decomposition of $X\in\mathfrak{g}$ is the unique decomposition $X=s+n$ with
- $s\in \mathfrak{g}$ semisimple (i.e. in the Lie algebra of a torus in $G$),
- $n\in\mathfrak{g}$ nilpotent (i.e. in the Lie algebra of a unipotent group in $G$), and
- $[s,n]=0$.
The commutation relation (3) is equivalent to $[s,X]=0$, and hence to $X\in\mathfrak{c}_{\mathfrak{g}}(s)$.
In very good characteristic, for every $X$, the stabilizer $C_G(X)$ is smooth and satisfies $\text{Lie}\, C_G(X) =\mathfrak{c}_{\mathfrak{g}}(X)$. Therefore $X\in \text{Lie}\, C_G(X)$, and replacing (3) by
3'. $C_G(X)(k)\subset C_G(s)(k)$
still uniquely specifies the Jordan-Chevalley decomposition. Does replacing (3) with (3') uniquely specify the Jordan-Chevalley decomposition in any characteristic?
Extra (more vague) questions if this is too hard:
Would looking at the centralizer instead of just $k$-points help?
Do centralizers of nilpotent elements (known if I understand correctly) give centralizers for general $X$ in an explicit enough way, or would verifying the question through the classification require ideas in addition to brute force?
