The Jordan-Chevalley Decomposition (proof)
We set so that is the direct sum of these subspaces, each of which is fixed by .
On the subspace , has the characteristic polynomial . What we want is a single polynomial such that
That is, has no constant term, and for each there is some such that
Thus, if we evaluate on the block we get .
To do this, we will make use of a result that usually comes up in number theory called the Chinese remainder theorem. Unfortunately, I didn’t have the foresight to cover number theory before Lie algebras, so I’ll just give the statement: any system of congruences — like the one above — where the moduli are relatively prime — as they are above, unless is an eigenvalue in which case just leave out the last congruence since we don’t need it — has a common solution, which is unique modulo the product of the separate moduli. For example, the system
has the solution , which is unique modulo . This is pretty straightforward to understand for integers, but it works as stated over any principal ideal domain — like — and, suitably generalized, over any commutative ring.
So anyway, such a exists, and it’s the we need to get the semisimple part of . Indeed, on any block differs from by stripping off any off-diagonal elements. Then we can just set and find . Any two polynomials in must commute — indeed we can simply calculate
Finally, if then so must any polynomial in , so the last assertion of the decomposition holds.
The only thing left is the uniqueness of the decomposition. Let’s say that is a different decomposition into a semisimple and a nilpotent part which commute with each other. Then we have , and all four of these endomorphisms commute with each other. But the left-hand side is semisimple — diagonalizable — but the right hand side is nilpotent, which means its only possible eigenvalue is zero. Thus and .