Exercise 10.19

Proof. If $\alpha \in {𝔽}^p$, then ${\alpha }^p=\alpha$, thus ${\alpha }^{p^k}=\alpha$ for all exponents $k\ge 0$.

In the extension ${𝔽}_{p^n}$ of ${𝔽}_p$, for all $\alpha \in {𝔽}_p$,

If the characteristic $p$ divides $n$, then $n=0$ in ${𝔽}_{p^n}$, thus $\mathrm{tr}(\alpha )=0$ for all $\alpha \in {𝔽}_p$.

Conversely, if $\mathrm{tr}(\alpha )=0$ for all $\alpha \in {𝔽}_p$, then $\mathrm{tr}(1)=n\cdot 1=0$, thus the characteristic $p$ divides $n$.

The extensions ${𝔽}_{p^n}$ of ${𝔽}_p$ that are such that the trace is identically zero on ${𝔽}_p$ are those which satisfy $p\mid n$. □