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Let $q$ be prime power, let $t \in \mathbb{N}$, and let $\omega \in \mathbb{F}_{q^t}*$. Consider the following invertible, $\mathbb{F}_q$-linear mapping on $\mathbb{F}_{q^t}$.

$g:\mathbb{F}_{q^t} \rightarrow \mathbb{F}_{q^t}$, $x \mapsto x^q\omega$

(The mapping $g$ is also $\mathbb{F}_{q^t}$-semilinear.) I believe (and would like to prove) that $g$ is cyclic as an element of ${\rm GL}_q(\mathbb{F}_{q^t})$, in the sense that there exists $v \in \mathbb{F}_{q^t}^*$ such that the elements

$v, vg, vg^2, \dots, vg^{t-1}$

form an $\mathbb{F}_q$-basis of $\mathbb{F}_{q^t}$. This is equivalent to saying that the minimal and the characteristic polynomial of $g$ coincide. Unfortunately, I have not yet found a proof (or a counterexample). I would appreciate any help/suggestions.

(Note that $g^t$ is given by $x \mapsto x \alpha$, where $\alpha = N_{\mathbb{F}_{q^t}:\mathbb{F}_q}(\omega)$ is the norm of $\omega$ over $\mathbb{F}_q$. Since $\alpha \in \mathbb{F}_q$, the minimal polynomial of $g^t$ is $x- \alpha$. Thus the minimal polynomial of $g$ divides $x^t-\alpha$. I would like to show that the minimal polynomial of $g$ is equal to $x^t-\alpha$.)

Thanks in advance!

sina
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2 Answers2

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Let $x \in \mathbb{F}_{q^t}$ be a primitive element. Then $x,x^q,x^{q^2},\ldots,x^{q^{t-1}}$ form an $\mathbb{F}_q$ basis of $\mathbb{F}_{q^t}$. That fact is well-known and you should find it in most books on finite fields; or you prove it yourself.

Now of course also $xw,x^qw,x^{q^2}w,\ldots$ form a basis.

So no need for characteristic polynomials here.

Dirk
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  • Thanks for your reply. This does not answer my question though, because $x, xg, xg^2, xg^3 ...$ equals $x, x^q\omega, x^{q^2}\omega^q\omega, x^{q^3}\omega^{q^2}\omega^q\omega$ and not $x\omega, x^q\omega, x^{q^2}\omega$. – sina Feb 02 '18 at 11:36
  • I'm afraid a primitive element does not necessarily yield a normal basis as you are implying. The smallest counterexample ($q=2$, $t=3$) is probably the case where $x$ is a zero of the polynomial $p(T)=T^3+T+1\in\Bbb{F}_2[T]$. Then $x$ has multiplicative order $7$, so it is a primitive element. The zeros of $p(T)$ are $x,x^2$ and $x^4$. But their sum is zero (look at the quadratic term of $p(T)$). – Jyrki Lahtonen Feb 03 '18 at 07:54
  • Of course (when $\omega=1$) we can still choose $x$ in such a way that $x,x^q.x^{q^2}$ et cetera form a basis. My point was that not every primitive element works in place of $x$. – Jyrki Lahtonen Feb 03 '18 at 08:05
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If the minimal polynomial of $g$, call it $m(T)=\sum_{i=0}^dm_iT^i$, had degree $d<t$ then every element of $\Bbb{F}_{q^t}$ would be a zero of the polynomial $$ \tilde{m}(x)=\sum_{i=0}^d m_ig^{[i]}(x), $$ where $g^{[i]}(x)$ stands for the $i$-fold composition $g\circ g\circ \cdots \circ g$. But, $\tilde{m}(x)$ has degree $q^d$, so cannot have $q^t$ zeros unless $d\ge t$.

This also settles your main claim. I got the impression that you figured that part out. It's similar to a standard proof of existence of a normal basis for the extension $\Bbb{F}_{q^t}/\Bbb{F}_q$, so I won't waste bandwidth by including those details.

Jyrki Lahtonen
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