additive character of a finite field, trace map to middle field The additive character of a finite field $\mathbb F_q$ is obtained by using the trace function to base field $F_p$.
Can we write some of them into a middle field. Using trace function from $\mathbb F_{q^n}$ to $\mathbb F_q$? If yes, how?
 A: Morgan gave you already the negative answer. For the sake of completeness...
There are plenty of trace maps around here. There is the relative trace
$$
tr^{q^n}_q:\Bbb{F}_{q^n}\to\Bbb{F}_q, x\mapsto x+x^q+x^{q^2}+\cdots+x^{q^{n-1}}.
$$
Then we have the two (absolute) trace maps of both these fields all the way down to the base field (assuming $q=p^m$ so $q^n=p^{mn}$:
$$
tr^q_p:\Bbb{F}_q\to\Bbb{F}_p,x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{m-1}}
$$
and
$$
tr^{q^n}_p:\Bbb{F}_q\to\Bbb{F}_p,x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{mn-1}}.
$$
It is a useful exercise to show that
$$
tr^{q^n}_p=tr^q_p\circ tr^{q^n}_q,
$$
i.e. composing the relative trace with the absolute trace of its codomain gives the absolute trace of the domain.
ALL of the additive characters of the finite field $\Bbb{F}_q$ can be gotten by the following recipe. The characters are parametrized by an element $a\in\Bbb{F}_q$, and are given by the formula
$$
\chi_a(x)=\zeta^{tr^q_p(ax)},
$$
where $\zeta=e^{2\pi i/p}$ is a primitive complex $p$th root of unity.
Proving this takes a while, but is easy if you already know that the number of additive characters must be equal to $q$. This is because it is easy to show that different choices of $a$ lead to different functions.
The choice $a=0$ gives the trivial character (=constant function taking the value $1$ everywhere). The character gotten with the choice $a=1$ is relatively often called the canonical character, but this term is not in universal use.
A: Not for an additive character; since a finite field has characteristic $p$, an additive map from $\mathbb{F}_{q^{n}} \to \mathbb{C}$ must send every element to a $p$th root of unity.  If you wrote into a middle field $\mathbb{F}_{q}$, how would you use these elements to obtain an additive map? I may be misunderstanding the question.
