Determining a dual of given basis Let $F :=\mathbb{F}_{16} = \mathbb{F}_2[x]/(x^4+x+1)$ and $K := \mathbb{F}_2$. Let $\{1=a^0,a,a^2,a^3\}$ be a basis in the vectorspace $F_K$. 

Find a (the?) dual of given basis.

By a dual basis we mean $\{b_0, b_1,b_2,b_3\}$ s.t
$$\mbox{Tr}_{F/K}(a_ib_j) =\delta _{ij} $$
where the trace $\mbox{Tr}_{F/K}(x) = x + x^2 + x^4 + x^8,\quad (x\in F)$.
I immediately have that
$$b_j + b_j^2 + b_j^4 + b_j^8 =\begin{cases}1, &j=0\\ 0, &j=1,2,3\end{cases} $$
and I could compute all the traces, but I don't see how to make progress with it.
An alternative idea is to use the fact that for every linear operator $T:F\to K$, there exists $c\in F$ with
$$\forall x\in F \left ( Tx = \mbox{Tr}_{F/K}(cx)\right ) $$
$1+a+a^2+a^3$ is a unique representation for some element in $F$, so if I define the "coordinate" functions
$$\pi _j(k_0a^0 + k_1a + k_2a^2 +k_3a^3) := k_j,\quad k_j\in K, j=0,1,2,3 $$
There is, for every $j$, a $b_j\in F: \pi_j(k_0a^0 + k_1a + k_2a^2 +k_3a^3) = \mbox{Tr}_{F/K}(b_j[k_0a^0 + k_1a + k_2a^2 +k_3a^3])$
It would turn out these elements $b_j$ are what I need to form a dual of the given basis, but it only reassures me such a basis exists. How would I find one explicitly?
 A: There is the following standard recipe that gives the dual of a monomial basis, i.e. a basis of the form $\mathcal{B}=\{1,a,a^2,\ldots,a^{n-1}\}$, where $F=K(a)$ is a separable degree $n$ extension of fields. Let $f(T)$ be the minimal polynomial of $a$ over $K$. Then do a polynomial division to get
$$
f(T)=(T-a)(c_0+c_1T+c_2+\cdots+c_{n-1}T^{n-1})
$$
with coefficients $c_i\in F$. Then the dual basis $\mathcal{B}$ of is
$$
\check{\mathcal{B}}=\{\frac{c_0}{f'(a)},\frac{c_1}{f'(a)},\ldots,\frac{c_{n-1}}{f'(a)}\}.
$$

Let's do an example with $K=\Bbb{F}_2$, $F=\Bbb{F}_{16}=\Bbb{F}_2(\gamma)$ with $\gamma^4+\gamma+1$. I refer you to this table listing elements of $F$ as powers of $\gamma$ (may make following the calculations below easier). Anyway, using $a=\gamma$, so $f(T)=T^4+T+1$ is the minimal polynomial. That's kinda dandy, because the derivative is then $f'(T)=4T^3+1=1$. All we need to do is that polynomial division:
$$
f(T)=(T+\gamma)(T^3+\gamma T^2+\gamma^2T+\gamma^3+1).
$$
So $c_0=\gamma^3+1$, $c_1=\gamma^2$, $c_2=\gamma$, $c_3=1$, and this is also the dual basis: $b_0=c_0$, $b_1=c_1$, $b_2=c_2$, $b_3=c_3$.
Let's do some checking. The zeros of $f(T)$ are $\gamma,\gamma^2,\gamma^4$ and $\gamma^8$. Their sum is the coefficient of $T^3$ in $f(T)$, i.e. $0$. This implies that $Tr(\gamma)=Tr(\gamma^2)=0$. Clearly $tr(1)=1+1+1+1=0$. Therefore we can conclude that $Tr(\gamma^3)=1$ for they cannot all vanish!
Verification of the dual basis property is then easy:
$$
\begin{aligned}
Tr(1\cdot b_0)&=Tr(\gamma^3+1)=1,\\ 
Tr(1\cdot b_1)&=Tr(\gamma^2)=0,\\
Tr(1\cdot b_2)&=Tr(\gamma)=0,\\ 
Tr(1\cdot b_3)&=Tr(1)=0\\
\\
Tr(\gamma\cdot b_0)&=Tr(\gamma^4+\gamma)=Tr(1)=0,\\
Tr(\gamma\cdot b_1)&=Tr(\gamma^3)=1,\\
Tr(\gamma\cdot b_2)&=Tr(\gamma^2)=0,\\
Tr(\gamma\cdot b_3)&=Tr(\gamma)=0,\\
\\
Tr(\gamma^2\cdot b_0)&=Tr(\gamma^5+\gamma^2)=Tr(\gamma)=0,\\
Tr(\gamma^2\cdot b_1)&=Tr(\gamma^4)=Tr(\gamma+1)=0+0=0,\\
Tr(\gamma^2\cdot b_2)&=Tr(\gamma^3)=1,\\
Tr(\gamma^2\cdot b_3)&=Tr(\gamma^2)=0,\\
\end{aligned}
$$
(leaving it to you to calculate the traces $Tr(\gamma^3b_i),i=0,1,2,3$). We see that the dual property holds as promised:
$$
Tr(\gamma^ib_j)=\begin{cases}1,&\ \text{if $i=j$, and}\\0,&\ \text{otherwise.}\end{cases}
$$

The same recipe for the dual basis of a monomial basis works for number fields. Therefore it may be easier to find a proof in some number theory text. The proof is not too difficult, but I don't remember all the details, and this answer is long enough as it is.

If you want to find the dual basis of a basis other than a monomial basis, then I cannot help you. Doing it the linear algebra way may be the best. The dual basis of a normal basis is always another normal basis, so if you are planning on using normal bases for representing elements of finite fields then there is some hope. I'm fairly sure there is a formula for that case as well.

Separability of the extension is absolutely necessary. Without separability the bilinear trace form will vanish, and the concept of a dual basis does not make any sense.
