The Jordan-Chevalley Decomposition (proof)
We now give the proof of the Jordan-Chevalley decomposition. We let have distinct eigenvalues
with multiplicities
, so the characteristic polynomial of
is
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
.
Very clear. Thank you.