Let $q < p $ be prime numbers. Let $r =ord_p(q)$ the minimal number $v$ s.t $q^v =_p 1$ (considering $q \in (\mathbb{Z}/p \mathbb{Z})^*$)
Let $T: \mathbb{F}_{q^r} \to \mathbb{F}_{q^r}$ be given by $T(X) = \sum_{i=0}^{r-1}X^{q^i}$.
In my book they claim that $T$ is surjective onto the subfield $\mathbb{F}_q$. Is this true? Are they saying that $Im(T) \subset \mathbb{F}_q$?
Also that $\mathbb{F}_{q^r}$ contains the $p$-$p^{th}$roots of unity and is generated by them as a vector field over $\mathbb{F}_q$.
Can you help me see why?
Yes, the trace is surjective. Note that $T(a)\in\Bbb F_q$ for $a\in \Bbb F_{q^r}$ (you can prove that by proving $T(a)^q=T(a)$). For each $b\in \Bbb F_q$, there are at most $q^{r-1}$ elements $a$ with $T(a)=b$ (as $T$ has degree $q^{r-1}$). This is only possible if for each $b\in \Bbb F_q$, there are exactly $q^{r-1}$ elements $a$ in $\Bbb F_{q^r}$ with $T(a)=b$. Thus $T$ is surjective.