2

Let $\sigma_p:\Bbb F_q \to \Bbb F_q$ be the Frobenius automorphism $\sigma_p(x)=x^p$ where $q=p^n$. Now viewing $V=\Bbb F_q$ as a vector space over $\Bbb F_p$ of dimension $n$.

Now how to prove that the matrix we get corresponding to Frobenius automorphism

  1. this matrix is diagonalizable over $\Bbb F_p$ iff $n|(p-1)$
  2. this matrix is diagonalizable over algebraic closure of $\Bbb F_p$ iff $(n,p)=1$

My try: I think the matrix of the linear transformation as the basis be $\{a,a^p,\cdots, a^{p^{n-1}}\}$. Then $$ \sigma_p(a)=a^p, \sigma_p(a^p)=a^{p^2}, \cdots, \sigma_p(a^{p^{n-1}})=a^{p^{n}}=a$$ where $V=\Bbb F_q=\Bbb F_p(a)$

$$(b_{ij})= \begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots& \ddots & \vdots &\vdots\\ 0 & 0 & \cdots & 1 & 0\\ \end{pmatrix}$$ is the matrix.

I can also calculate the characteristic polynomial of $(b_{ij})=x^n+(-1)^{2n-1}=X^n-1$= minimal polynomial [the minimal polynomial has to divide X^n -1 It cannot be a proper divisor cause otherwise, min polynomial would have deg less than $n$ Then the extension deg would be less than $n$]

Diagonalizable over F_p iff it splits into distinct linear factors over F_p iff F_p has an element of multiplicative order $n$ iff n divides the order of the cyclic multiplicative group, which is p-1.

How to prove that 2) "this matrix is diagonalizable over algebraic closure of $\Bbb F_p$ iff $(n,p)=1$"?

Ri-Li
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  • A couple of things to check first. Your "basis" seems to have $n+1$ elements. And your characteristic polynomial seems not to be of degree $n$. – ancient mathematician Apr 30 '19 at 06:39
  • Hi, yes I am sorry. Is that not a basis then? I think I have done it wrong. Can you give a solution from scratch? – Ri-Li Apr 30 '19 at 06:59
  • I think you should repair the question first. Tell us what $a$ is. Then isn't it the case that omitting $1$ gives a basis? Once that's done you'll have $m_B(X)|\chi_B(X)=X^n-1$ and the rest follows from the standard criterion for diagonalisability. – ancient mathematician Apr 30 '19 at 07:08
  • I edited the question, from the part where you get $a$ that is my try. Let me highlight that... – Ri-Li Apr 30 '19 at 07:15
  • But you have agreed that the basis is wrong, you need to change that. After all the sum of the conjugates of $a$ is fixed by $\sigma$ and so is a multiple of $1$. Delete the $1$ and then work out the characteristic polynomial properly. To get a suitable $a$ you need the Normal Basis Theorem. – ancient mathematician Apr 30 '19 at 08:05
  • Is it true now? – Ri-Li Apr 30 '19 at 09:16
  • @ancientmathematician help me now!! – Ri-Li Apr 30 '19 at 10:02
  • Surely the characteristic polynomial is $X^n-1$. Now over $\mathbb{F_p}$ the matrix diagonalises iff $m_B$ is a product of distinct linear factors, and this happens precisely when $n|(p-1)$. Similarly we can deal with the algebraic closure case. – ancient mathematician Apr 30 '19 at 12:26
  • Can you write in details? – Ri-Li Apr 30 '19 at 18:47
  • +1 to all. A related older thread. Not much more information there, and you worked harder at it than the previous asker. – Jyrki Lahtonen May 01 '19 at 13:02

1 Answers1

2

For 2). Since $B$ is a Frobenius block, each of its eigenspaces has dimension $1$.

Thus $B$ is diag. over $\overline{F_p}$ iff each of the (non-zero) roots $x_i$ of $x^n-1$ has multiplicity $1$ iff $n{x_i}^{n-1}\not= 0$ iff $(n,p)=1$.

  • I haven't downvoted you by the way. I am trying to understand your answer and in the dilemma whether it is right or wrong!! xD – Ri-Li May 01 '19 at 00:47