6

Prove that over finite field $\mathbb F$ exists additive Jordan-Chevalley decomposition: for all matrix $M$ there are semisimple matrix $M_{s}$ and nilpotent matrix $M_{n}$ such that $M=M_{s}+M_{n}$.

I proved that $M^n=M^k$ for some $n,k \in \mathbb N$ as $\mathbb F$ is finite. Then considering $M$ is inversible we get that $M^{n-k}=E$. But, unfortunetly, polynomial $x^{n-k} - 1$ is not separable if $p|(n-k)$. If $n-k$ is not even we have decomposition $M=E + (M-E)$. In case $n-k$ is even $M-E$ is not nilpotent. And i got stucked.

Also I proved it in cases characteristic polynomial $f$ of $M$:

  1. all roots of $f$ belong $\mathbb F$

  2. $f=gh$, where all roots of $g$ belongs $\mathbb F$ and $h$ is irreducible.

Rodrigo de Azevedo
  • 1
SaveMyLife
  • 149

1 Answers1

1

Hint : Since you have shown that the Jordan decomposition exists when $f$ splits in $\mathbb{F}$, the same proof you found should work when $f$ splits over a field $K$. So you have a Jordan decomposition of $M$ on $K$, where $K$ is the splitting field of $f$. Then use the fact that since $\mathbb{F}$ is perfect, you can be sure that $K$ is a Galois extension of $\mathbb{F}$. Make the Galois group of $K$ on $\mathbb{F}$ acts on this decomposition to deduce that the matrices actually belong to $\mathbb{F}$.

brunoh
  • 2,436