Over a perfect field, every endomorphism has a unique commuting semisimple-plus-nilpotent decomposition, polynomial in the endomorphism

Choose an algebraic closure $\mathrm{\Omega }\mathrm{/}F$ as in An algebraic closure of a field. Let $q$ be the product of the distinct monic irreducible factors of the characteristic polynomial of $T$, and let $E\subseteq \mathrm{\Omega }$ be its splitting field. By [L1], the polynomial $q$ is separable, so [L4] makes $E\mathrm{/}F$ finite Galois with fixed field $E^{\mathrm{Gal}(E\mathrm{/}F)}=F$. The characteristic polynomial splits over $E$, and [L2] gives Jordan form for the matrix of $T$ over $E$. Thus $E^n$ decomposes as the direct sum of generalized eigenspaces $G_{\lambda }$, and $(T_E-\lambda I){\mathrm{\mid }}_{G_{\lambda }}$ is nilpotent.

Let $P_{\lambda }$ be the projection onto $G_{\lambda }$ along the sum of the other generalized eigenspaces. By [L3], each $P_{\lambda }$ is a polynomial in $T_E$. Define $S_E:={\sum }_{\lambda }\lambda P_{\lambda }$ and $N_E:=T_E-S_E={\sum }_{\lambda }(T_E-\lambda I)P_{\lambda }$. Then $T_E=S_E+N_E$, the operators commute because they are polynomials in $T_E$, the minimal polynomial of $S_E$ divides ${\prod }_{\lambda }(x-\lambda )$ and therefore has distinct roots, so [L5] makes $S_E$ semisimple, and $N_E$ is nilpotent because its restriction to each $G_{\lambda }$ is $(T_E-\lambda I){\mathrm{\mid }}_{G_{\lambda }}$.

Suppose also that $T_E=S_E^{\mathrm{\prime }}+N_E^{\mathrm{\prime }}$ with $S_E^{\mathrm{\prime }}$ semisimple, $N_E^{\mathrm{\prime }}$ nilpotent, and $S_E^{\mathrm{\prime }}{N}_E^{\mathrm{\prime }}=N_E^{\mathrm{\prime }}{S}_E^{\mathrm{\prime }}$. Because $N_E^{\mathrm{\prime }}$ commutes with $T_E$, every generalized eigenspace $G_{\lambda }$ from step 1.1 is invariant under both $S_E^{\mathrm{\prime }}$ and $N_E^{\mathrm{\prime }}$. On $G_{\lambda }$, the operator $T_E$ has only the eigenvalue $\lambda$, while $N_E^{\mathrm{\prime }}$ has only the eigenvalue $0$; hence the semisimple operator $S_E^{\mathrm{\prime }}{\mathrm{\mid }}_{G_{\lambda }}$ has only the eigenvalue $\lambda$, so $S_E^{\mathrm{\prime }}=\lambda I$ on $G_{\lambda }$ and therefore $N_E^{\mathrm{\prime }}=T_E-\lambda I$ there. Thus $S_E^{\mathrm{\prime }}=S_E$ and $N_E^{\mathrm{\prime }}=N_E$.

Let $\sigma \in \mathrm{Gal}(E\mathrm{/}F)$. Acting entrywise, $\sigma$ fixes the matrix of $T_E$ and permutes the roots of its characteristic polynomial. Applying $\sigma$ to the construction in step 2.1 therefore produces another commuting semisimple-plus-nilpotent decomposition of $T_E$. Uniqueness in step 3.1 forces $\sigma (S_E)=S_E$ and $\sigma (N_E)=N_E$.

Choose an $F$-basis $B_1,\dots ,B_d$ of the finite-dimensional algebra $F[T]$. The same matrices form an $E$-basis of $E[T_E]$. Write $S_E={\sum }_i{c}_i{B}_i$ with $c_i\in E$. Step 4.1 and uniqueness of coordinates give $\sigma (c_i)=c_i$ for every $\sigma \in \mathrm{Gal}(E\mathrm{/}F)$, so [L4] gives $c_i\in F$. Hence $S_E$ is the scalar extension of a polynomial $T_s=p(T)$ with $p\in F[x]$. The same argument applies to $N_E$, giving $T_n=q(T)$ with $q\in F[x]$. Extending the identities of step 2.1 back down to $F$ gives $T=T_s+T_n$ and $T_s{T}_n=T_n{T}_s$; semisimplicity and nilpotence descend because their minimal-polynomial identities have coefficients in $F$. Uniqueness follows after extension to $E$ from step 3.1.