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Let $q$ be a prime power. Denote by $\mathbb{F}_{q^m}$ the finite field with $q^m$ elements for any positive integer $m$. Fix a positive integer $r$, and fix $\alpha \in\mathbb{F}_{q^r} $. Prove the trace $T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\alpha)=0$ if and only if there exists $\beta \in \mathbb{F}_{q}$ such that $\alpha=\beta^q-\beta$.

Things I know:

We have that $[\mathbb{F}_{q^r}: \mathbb{F}_{q}]=r$. Then if $d = [\mathbb{F}_{q}(\alpha):\mathbb{F}_{q}]$, we have that $d|r$.

By definition we have $T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\alpha) = \sum_{\sigma} \sigma (\alpha)$.

So for $(\leftarrow)$ of the proof,

$T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\alpha)= T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\beta^q-\beta)=T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\beta^q) - T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\beta)$. We just need to show $T_{\mathbb{F}_{q^r}/ \mathbb{F}_q}(\beta^q)=0$. But I'm stuck on this, as I think I'm missing some facts on trace to show that.

For $(\rightarrow)$ of the proof, I'm not sure how to proceed.

Any help will be appreciated!

Korn
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    Are you sure you mean $\beta \in \mathbb F_q$ and not $\beta \in \mathbb F_{q^r}$? – Arkady Jun 23 '22 at 19:34
  • In my humble opinion your question is a duplicate of this older version. Because I answered that oldie, I will, for the time being, not cast a vote. I have no less than four applicable dupehammers, so my vote would take immediate effect. If you agree with me, do @-ping me. – Jyrki Lahtonen Jun 23 '22 at 20:00
  • @JyrkiLahtonen My apologies. I should have checked before answering this question. I will let OP make a decision on this. – Arkady Jun 23 '22 at 20:45

1 Answers1

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We have $$Tr_{q^r|q}(\alpha)=\alpha+\alpha^q...+\alpha^{q^{r-1}}\in \mathbb{F}_q.$$ If $\alpha=\beta^q-\beta$ for some $\beta\in \mathbb{F}_{q^r}$, then $$Tr_{q^r|q}(\alpha)=Tr_{q^r|q}(\beta)^q-Tr_{q^r|q}(\beta)=0.$$ If $$Tr_{q^r|q}(\alpha)=0,$$ then let $\beta$ be a root of $x^q-x-\alpha$ in some extension of $\mathbb{F}_q$. Then, $\beta^q-\beta=\alpha$ and $$\begin{align*} 0=Tr_{q^r|q}(\alpha)&=Tr_{q^r|q}(\beta)^q-Tr_{q^r|q}(\beta)\\ &=\beta^{q^r}-\beta. \end{align*}$$ Hence, $\beta^{q^r}=\beta$ and $\beta\in\mathbb{F}_{q^r}$.

Yano
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