# Frobenius Morphism

• March 13th 2011, 10:18 PM
h2osprey
Frobenius Morphism
Let $\mathbb{F}$ be a finite field of characteristic $p.$ As such it is a finite-dimensional vector space over $\mathbb{Z}_p$.

a) Prove that the Frobenius morphism $Fr:\mathbb{F} \rightarrow \mathbb{F}, Fr(a) = a^p$ is a linear map over $\mathbb{Z}_p.$
b) Prove that the geometric multiplicity of 1 as an eigenvalue of $Fr$ is 1.
c) Let $\mathbb{F}$ have dimension 2 over $\mathbb{Z}_7.$ Prove that 2 is not an eigenvalue of $Fr$.

I solved a) using Fermat's Little Theorem, but am unsure about b) and c). I think I solved c) by considering $Fr^7$, but I'm not sure if there's a simpler proof.

Thanks!
• March 14th 2011, 04:34 AM
tonio
Quote:

Originally Posted by h2osprey
Let $\mathbb{F}$ be a finite field of characteristic $p.$ As such it is a finite-dimensional vector space over $\mathbb{Z}_p$.

a) Prove that the Frobenius morphism $Fr:\mathbb{F} \rightarrow \mathbb{F}, Fr(a) = a^p$ is a linear map over $\mathbb{Z}_p.$
b) Prove that the geometric multiplicity of 1 as an eigenvalue of $Fr$ is 1.

This means you have to prove the dimension of the eigenspace corresponding to 1 is

1, but $a=:Fr(a)=a^p\Longleftrightarrow a=a^p\Longleftrightarrow a\in \mathbb{Z}_p$ , so...

c) Let $\mathbb{F}$ have dimension 2 over $\mathbb{Z}_7.$ Prove that 2 is not an eigenvalue of $Fr$.

2 is an eigenvalue iff $\exists\,0\neq a\in\mathbb{F}\,\,s.t.\,\,a^7=Fr(a)=2a\Longleftrig htarrow a^6=2$ , which

of course is impossible.

Tonio

I solved a) using Fermat's Little Theorem, but am unsure about b) and c). I think I solved c) by considering $Fr^7$, but I'm not sure if there's a simpler proof.

Thanks!

.
• March 14th 2011, 06:28 PM
h2osprey
Quote:

Originally Posted by tonio
.

I get b) now, but for c) why is $a^6 = 2$ not possible?