Let $J$ be a matrix with entries in some field $\mathbb{k}$ that is in Jordan normal form. More explicitely, $$ J = \begin{pmatrix} J(\lambda_1, n_1) & & \\ & \ddots & \\ & & J(\lambda_r, n_r) \end{pmatrix} $$ where $$ J(\lambda, n) = \begin{pmatrix} \lambda & 1 & & \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix} \,. $$ The matrix $J$ admits a Jordan–Chevalley decomposition $J = D + N$. The diagonalizable part $D$ is given by the diagonal of $J$, and the nilpotent part $N$ is given by the strictly upper triangular part of $J$. It follows from the abstract construction of the Jordan–Chevalley decomposition (via the Chinese remainder theorem) that there exist polynomials $p$ and $q$ in $\mathbb{k}[x]$ with $D = p(J)$ and $N = q(J)$. One can furthermore assume that both $p$ and $q$ have no constant part.
Question. How can the existence of the polynomials $p$ and $q$ be derived directly from the Jordan normal form?
(Here “directly” entails “without use of the Chinese remainder theorem”.)
The motivation behind this question is the observation that there seem to be two approaches to showing the existence of the Jordan–Chavalley decomposition of a matrix $A$ (if $\mathbb{k}$ is algebraically closed):
One can conjugate $A$ to a matrix $J$ that is Jordan normal form. The decomposition of $J$ into a diagonal part and a strictly upper triangular part corresponds to the desired decomposition for $A$.
One can consider the characteristic polynomial $\chi$ of $A$ (or any other nonzero polynomial that annihilates $A$) and apply the Chinese remainder theorem to $\mathbb{k}[x] / (\chi)$ to construct the polynomial $p$, and then define $q$ as $1 - p$.
The first approach only needs the existence of the Jordan normal form and explains explicitely how the Jordan normal form of $A$ relats to the Jordan–Chevalley decomposition of $A$. But it seem to have the disadvantage that it does not show the existence of the polynomials $p$ and $q$. Hence the question.