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My objective is to find the Galois conjugates of elements in $\mathbb{F}_{p^n}$ where $p$ is a prime and $n > 1$. It seems that the Frobenius automorphism is what I should be looking into.

Section 1.2 of The rank one stark conjecture for abelian extensions of quadratic imaginary number fields states:

let $p$ be rational prime and consider the finite field $\mathbb{F}_{p^n}$ with $p^n$ elements. Recall that $\mathbb{F}_{p^n}$ is Galois over its prime subfield $\mathbb{F}_{p}$ and that $\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p})$ is cyclic. It is generated by the automorphism $\sigma_p \, : \, \mathbb{F}_{p^n} \rightarrow \mathbb{F}_{p^n}$ defined by $x \mapsto x^p$. This map $\sigma_p$ is called the Frobenius automorphism or the Frobenius map.

Furthermore, section 4.1 of Permutation code equivalence is not harder than graph isomorphism when hulls are trivial states:

Consider $\mathbb{F} = \mathbb{F}_{p^m}$ with $m > 1$ and $p$ is a prime number. The idea is to consider a Hermitian inner product $\langle \textbf{$x$}, \, \textbf{$y$} \rangle_\theta \overset{\mathrm{\triangle}}{=} \Sigma^{n}_{i=1} x_i \theta(y_i)$ where $\theta$ is an automorphism of $\mathbb{F}_{p^m}$. We recall that such automorphisms are in the form of $z \mapsto z^{p^e}$ where $e$ lies in $[\![0, m - 1]\!]$.

We can see that both automorphisms $\sigma_p$ and $\theta$ differ by the exponent $e$. My questions are:

  1. How is $e$ chosen in the second definition as there are $m$ possible choices?
  2. Is there a way to link the two definitions concluding that Hermitian inner product's is indeed Frobenius automorphism?

In the case where it helps to explain with an example, the field $\mathbb{F}_{2^2}$ can be used :).

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    $\begingroup$ Can you know what exactly are $\;\alpha,\,1+\alpha\;$ ? Because if you don't know this then it'll be hard to help you. This is the very first I'd check and write down in the very body of the question. $\endgroup$
    – DonAntonio
    Commented May 22, 2021 at 9:42
  • $\begingroup$ The definition you refer to is for the conjugate elements [so could be more than one] of a given element in a field extension. Note early in tht link it says an element is considered conjugate to itself. There may be others... $\endgroup$
    – coffeemath
    Commented May 22, 2021 at 9:43

3 Answers 3

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There are the following three classes of conjugates:

$\{0\}$ (the roots of the minimal polynomial $X$)

$\{1\}$ (the roots of the minimal polynomial $X+1$)

$\{\alpha, \alpha+1\}$ (the roots of the minimal polynomial $X^2 + X + 1$)

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Since you are working in a finite field, you should perhaps try and apply the most general tool available in Galois theory of finite fields, which is the Frobenius automorphism.

In your special case, consider the quadratic extension of finite fields $F_4/F_2$, with cyclic Galois group $<\phi>$ of order 2, where the Frobenius automorphism is defined by $\phi (x)=x^2$ (2 is the cardinal of the base field). In particular, if $F_4= F_2 (\alpha)$, the orbit of $\alpha$ under the action of $\phi$ consists only of $\alpha, \alpha^2$ because $\alpha^4 = \alpha$ since ${F_4}^*$ has order 3. But $\alpha^3 = 1$ also implies $\alpha^2 = \alpha +1$ in characteristic 2. Finally, the 4 elements of $F_4$ are $0,1,\alpha,\alpha +1$ as desired.

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In this very small case, we know that $0$ and $1$ are self-conjugate, and that there are exactly two other elements, and that those other two elements are conjugate, however else we describe them. And for any $\alpha\not\in\{0,1\}$, likewise $\alpha+1\not\in\{0,1\}$ and is distinct from $\alpha$, so $\alpha$ and $\alpha+1$ must be mutual conjugates.

Yes, it is illuminating to consider the irreducible polynomials, etc., but is not strictly necessary. Nor is it necessary to choose a way to specify/constract the $\alpha\in\{0,1\}$, if you grant that at least one exists.

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