Let $V$ be a finite dimensional vector space over an algebraically closed field $k$, and let $a$ be an endomorphism of $V$.
Theorem:
(i) There exist unique endomorphisms $a_s, a_n$ of $V$ such that $a_s$ is semisimple, $a_n$ is nilpotent, $a_s a_n = a_n a_s$, and $a = a_s + a_n$.
(ii): $a_s$ and $a_n$ are polynomials in $a$.
I understand most of the proof I'm reading (in Springer, Linear Algebraic Groups), except the uniqueness statement. What Springer does is define $a_s$ and $a_n$ to be polynomials in $a$ which satisfy the conditions in (i). He then says: "To prove (uniqueness), let $a = b_s + b_n$ be a second decomposition with the properties of (i). From (ii) we infer that $b_s$ and $b_n$ commute with $a_s$ and $a_n$."
There is a problem here though. He only proved that the specific $a_s, a_n$ he constructed to satisfy (i) were polynomials in $a$. He did not show that any such $a_s,a_n$ had to be polynomials. How can one work around this?
As usual, my difficulty with understanding the proof can be resolved in a couple of sentences. It doesn't matter if we don't yet know that $b_s$ and $b_n$ are polynomials in $a$. The point is that $a_s$ and $a_n$ are polynomials in $a$. Therefore the fact that $b_s$ and $b_n$ commute with $a$ implies that they also commute with $a_n$ and $a_s$.