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Let $\alpha$ be a root of $f(x) : = x^4 + x + 2 \in \mathbb{F}_{3}[x]$. One verifies that $f$ is irreducible and so $\mathbb{F}(\alpha)/\mathbb{F}$ is Galois with Galois group, $G$, generated by the Frobenius $\varphi : \mathbb{F}_{3}(\alpha) \rightarrow \mathbb{F}_{3}(\alpha)$. By the Fundamental Theorem of Galois Theory $\mathbb{F}_{9} \subseteq \mathbb{F}_{3}(\alpha)$ is the fixed field of $\varphi^2$ and I would like to:

  1. Find a primitive element for $\mathbb{F}_{9}$ expressed in terms of $\alpha$.
  2. Find the minimal polynomial for this primitive element over $\mathbb{F}_{3}$, which is necessarily a quadratic.

My approach was as follows:

$\{ 1 , \alpha , \alpha^2 , \alpha^3 \}$ is a $\mathbb{F}_{3}$-basis for $\mathbb{F}_{3}(\alpha)$ and for arbitrary $\beta : = b_{0} + b_{1} \alpha + b_{2} \alpha^2 + b_{3} \alpha^3 \in \mathbb{F}_{3}(\alpha)$, $\varphi^2(\beta) = \beta$ if and only if $$ b_{0} + b_{1} \alpha + b_{2} \alpha^2 + b_{3} \alpha^3 = b_{0} + b_{1} \alpha^9 + b_{2} \alpha^{18} + b_{3} \alpha^{27}. $$ After using the fact that $\alpha$ satisfies the relation $\alpha^4 = 2 \alpha + 1$ to simplify the right-hand side of the equation above followed by solving a system of equations I was able to deduce that $\beta$ is necessarily of the form $\beta = a + b\alpha + b \alpha^2$, where $a , b\in \mathbb{F}_{3}$.

Thus $\mathbb{F}_{9} = \{ a + b\alpha + b \alpha^2 :a , b \in \mathbb{F}_{3} \}$ and any $a + b \alpha + b \alpha^2$ with $b \neq 0$ should be a primitive element for the extension $\mathbb{F}_{9}/\mathbb{F}_{3}$. However, when I try $\beta : = \alpha + \alpha^2$, we see that $\beta^2 = 1 + 2 \alpha + \alpha^2 + 2 \alpha^3$, which leads me to believe that $\beta$ will not satisfy a degree 2 polynomial because I will be unable to kill off that $\alpha^3$ term.

Can anyone point out what is going wrong or where I may be mistaken?

user26857
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Oiler
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  • Which definition of a primitive element are you working with? In the case of finite fields it is standard to reserve the term for generators of the multiplicative group. Elsewhere in the field theory $\beta\in L$ would be called primitive whenever $L=K(\beta)$, but in finite fields we usually require more. – Jyrki Lahtonen Jul 01 '21 at 09:04
  • See here for a more detailed explanation. – Jyrki Lahtonen Jul 01 '21 at 09:06
  • @JyrkiLahtonen I meant primitive in the sense of the primitive element theorem, i.e. an element one adjoins to the base field to yield the field extension. – Oiler Jul 01 '21 at 09:19

1 Answers1

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Your mistake comes somewhere in your simplification of $b_0+b_1\alpha^9+b_2\alpha^{18}+b_3\alpha^{27}$ which you haven't shown us.

Quicker method: Recall the trace map for finite separable field extensions are surjective. In this case, $\operatorname{Tr}_{\mathbb{F}_{81}/\mathbb{F}_9}\colon\mathbb{F}_{81}\to\mathbb{F}_9$ is $y\mapsto y+\varphi^2(y)$, so each elements of $\mathbb{F}_9$ can be written as $y+\varphi^2(y)$ for some $y\in\mathbb{F}_{81}$ hence you just have to make sure that $y$ doesn't land you in $\mathbb{F}_3$.

So let's try the simplest case $y=\alpha$, and check $\alpha+\varphi^2(\alpha)=\alpha+\alpha^9=2\alpha+\alpha^2+\alpha^3$ which is clearly not in $\mathbb{F}_3$.

Now we check $\beta=2\alpha+\alpha^2+\alpha^3$, $\beta^2=\alpha^2+\alpha^3+2\alpha^4+2\alpha^5+\alpha^6=\alpha^2+\alpha^3+(1+2\alpha)(2+2\alpha+\alpha^2)=2=-1$ giving the min poly $x^2+1$. If you want generators of $\mathbb{F}_9^\times$ instead then just make $1+\beta$ instead.

user10354138
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