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Let $q$ be a prime power and $n$ be a positive integer. Let $U$ be a $\mathbb{F}_q$-vector subspace of the finite field $\mathbb{F}_{q^n}$ of dimension $n-1$. Let $V$ be the kernel of the trace map $Tr: \mathbb{F}_{q^n} \rightarrow \mathbb{F}_q$. It is known that $V$ is also a vector subspace of dimension $n-1$. Then can we say that $U=aV$ for some non-zero $a\in\mathbb{F}_{q^n}$?

jimm
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1 Answers1

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Yes, simply say that the $x\mapsto Tr(ax)$ give $q^n$ distinct $\Bbb{F}_q$-linear maps $\Bbb{F}_{q^n}\to \Bbb{F}_q$, so they represent all of them.

Take a $\Bbb{F}_q$-linear isomorphism $h:\Bbb{F}_{q^n}/U \to \Bbb{F}_q$, then $h = Tr(a.)$ for some $a\in \Bbb{F}_{q^n}^*$ so $U=\ker h =\ker(Tr(a.))=a^{-1} \ker(Tr)$

reuns
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