how to find generator in $ F_2[x]/(f(x))$? a finite field $F_{2^n}\cong F_2[x]/(f(x))$($f(x)$is irreducible polynomial with the degree of $n$), so the elements in $F_{2^n}$ can be seen a polynomial modular $f(x)$, that is :$$\{g_0(x),g_1(x),...,g_{2^n-1}(x)\}_{f(x)}$$
There is a  isomorphism  that $\varphi=\left\{ \begin{array}{l l} F_{2^n}\to F_2[x]/(f(x)) \\ \xi \to k(x) & \quad \text{ $\xi$ is primitive element}\end{array} \right.\  $
Because $\xi$ is primitive element in $F_{2^n}$, so $\varphi(\xi)=k(x)$ can also generate all the elements in $ F_2[x]/(f(x))$.
My question is that how to  find $k(x)$, and what the structure of $k(x)$? 
thanks a lot
 A: The answer to this question depends on the representation of $\mathbb F_{2^n}$
you have in mind.
Suppose the elements of $\mathbb F_{2^n}$ just have names e.g.
$a_0, a_1, \ldots, a_{2^n-1}$, without any meaning ascribed to
them, and you use $2^n\times 2^n$ addition 
and multiplication tables for arithmetic in $\mathbb F_{2^n}$.
You also have a more concrete representation of $\mathbb F_{2^n}$
as $\mathbb F_{2}[x]/(f(x))$, in which representation,
addition and multiplication tables are not necessary
since arithmetic on the $g_i(x)$'s is done as polynomial
addition in $\mathbb F_{2}[x]$ and polynomial multiplication in
$\mathbb F_{2}[x]$ followed by a residue computation modulo $f(x)$.
In this case, if you know that $a_i \in \mathbb F_{2^n}$
is a zero of $f(x)$, then the desired isomorphism is
$$a_i \leftrightarrow x$$
and the images of all other $a_j$ follow from this. For example,
if $a_j = (a_i)^2$ (which we need to use the tables to figure
out), then
$$a_j \leftrightarrow x^2$$
and if $a_i+a_j = a_i + (a_i)^2 = a_k$, then
$$a_k \leftrightarrow x + x^2$$
and so on and so forth.
But what if you do not know which of the $a_m$ are zeroes
of $f(x)$?  Well, the brute-force way is simply to try
each $a_m$ by evaluating $f(a_m)$ (via the addition
and multiplication tables) and checking if the evaluation
results in $0$. A slightly more efficient way is to not
bother evaluating  any of $f(a_m^2)$, $f(a_m^{2^2})$, $\cdots$, 
$f(a_m^{2^{n-1}})$
if $f(a_m) \neq 0$ because of $a_m$ is not a zero of
$f(x)$, then its conjugates cannot be zeroes of $f(x)$ either.
