Theorem 1

Let $E$ be a finite dimensional vector space over a commutative field $K$ and let $u$ be an endomorphism of $E$. Then $u$ has a Jordan decomposition $(u_s, u_n)$ if and only if the eigenvalues of $u$ are separable over $K$. Moreover the decomposition is unique, the characteristic polynomials of $u$ and $u_s$ coincide, and there exist polynomials $P, Q \in K[X]$, with no constant terms, such that $u_s = P(u)$ and $u_n = Q(u)$.

A) Let us first of all prove the following special case