- #1

mathmari

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Let $q$ be a power of a prime and $n\in \mathbb{N}$. We symbolize with $Tr$ the map of the trace from $\mathbb{F}_{q^n}$ to $\mathbb{F}_q$, i.e. $Tr:\mathbb{F}_{q^n}\rightarrow \mathbb{F}_q$, $\displaystyle{Tr(a)=\sum_{j=0}^{n-1}a^{q^j}}$. I want to calculate the dimension of the image of the linear map $\theta : \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n}$, $\theta (\beta)=\beta^q-\beta$.

We have that $\dim (\mathbb{F}_{q^n})=\dim (\ker (\theta))+\dim (\text{im} (\theta))$, right?

The dimension of $\mathbb{F}_{q^n}$ is $q^n$, isn't it?

The kernel of the linear map is $$\ker (\theta)=\{\beta \in \mathbb{F}_{q^n}: \theta (\beta )=0\}=\{\beta \in \mathbb{F}_{q^n}: \beta^q-\beta=0\}$$ How many elements does this set have? (Wondering)

After that I want to show that $\ker (Tr)=\{\beta^q-\beta :\beta \in \mathbb{F}_{q^n}\}$.

We have that $$\ker (Tr)=\{a\in \mathbb{F}_{q^n} : Tr(a)=0\}=\left \{a\in \mathbb{F}_{q^n} : \sum_{j=0}^{n-1}a^{q^j}=0\right \}$$ Could you give me a hint how we could continue? (Wondering)