Let $\mathbb{F}_{9} \cong \mathbb{F}_{3} / (X^{2}+1)$ be a finite field with nine elements and $\sigma_{9}: \mathbb{F}_{9} \rightarrow \mathbb{F}_{9}$ the Frobenius automorphism.

Has the field a decomposition with $M_{x}=\{ \sigma_{9}^{i}(x) : i \in \mathbb{Z} \ and\ x \in \mathbb{F}_{9} \} \subset \mathbb{F}_{9}$, s.t. : $\mathbb{F}_{9}= \cup_{x\in \mathbb{F}_{9}} M_{x}$ ?

To my mind it's the identity, s.t. #$M_{x}=1, \forall x \in \mathbb{F}_{9}$ and #$\{M_{x} : x \in \mathbb{F}_{9} \}=9$, because the frobenius map is an automorphism.


I think that your mistake is in what the frobenius map does. It is true that $x^9 = x$ for all $x \in \mathbb{F}_9$. However, $\sigma_9(x) = x^3$. This is not the identity map.

The sets $M_x$ partition $\mathbb{F}_9$ into equivalence classes consisting of elements who have the same minimal polynomial over $\mathbb{F}_3$.

The elements of $\mathbb{F}_9$ are the roots of all irreducible polynomials of degree $1$ and $2$ over $\mathbb{F}_3$. These polynomials are: $$x, x-1, x-2, x^2 + 1, x^2+ x + 2, x^2 + 2x + 2.$$

Let $\alpha \in \mathbb{F}_9$ be such that $\alpha^2 + 1 = 0$. Let's compute the sets $M_x$.

\begin{array}{|c | c|} \hline M_0 & 0 \\ M_1 & 1 \\ M_2 & 2 \\ M_\alpha & \alpha, 2\alpha \\ M_{\alpha + 1} & \alpha + 1, 2\alpha + 1 \\ M_{\alpha + 2} & \alpha + 2, 2\alpha + 2 \\ \hline \end{array}

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    $\begingroup$ And it is always true that if $K/F$ is a normal field extension then the set $M_x = \{y \in K, y = \sigma(x), \sigma \in Aut(K/F)\}$ are the roots of the minimal polynomial of $x$ over $F$. In finite fields, the Frobenius automorphism generates the Galois group, and what you wrote follows. $\endgroup$ – reuns Dec 11 '16 at 0:04
  • $\begingroup$ $\sigma_9$ has order $2$ so its orbits have length $1$ or $2$. Galois theory says that the elements $x$ have orbit size $1 \iff x \in \Bbb{F}_3$. The other six elements have orbits size $2$. $\endgroup$ – Marc Bogaerts Dec 12 '16 at 14:16

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