Calculate the trace of all elements in $F_8$ I got the following exercise where you have to calc the trace of all elements in ${F_8}$ which is constructed as ${F_2}[x]$/(${x^3+x+1}$)${F_2}[x]$.
Up to now I did those steps:
1) Find all elements in ${F_8}$ which are in my opinion: $0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1$
2) Then I found those traces for the elements:
$Tr(0)=0\\
Tr(1)=1\\
Tr(x)=x+x^2+x^4\\
Tr(x+1)=Tr(x)+Tr(1)=x^4+x^2+x+1\\
Tr(x^2)=x^8+x^4+x^2 \text{(must this be reduced!?)}\\
Tr(x^2+1)=Tr(x^2)+Tr(1)=x^8+x^4+x^2+1\\
Tr(x^2+x)=Tr(x^2)+Tr(x)=x^8+x\\
Tr(x^2+x+1)=Tr(x^2)+Tr(x)+Tr(1)=x^8+x+1$
Is this procedure correct? Thank you!
 A: We have here an extension of dimension three of  the prime field $\;\Bbb F_2\;$ , and in fact
$$\Bbb F_{2^3}=\Bbb F_8\cong\Bbb F_2[x]/\langle x^3+x+1\rangle = \text{Span}_{\Bbb F_2}\{\overline 1,\overline x,\overline{x^2}\}\;$$
where for $\;v\in\Bbb F_2[x]\;,\;\;\overline v:=v+\langle x^3+x+1\rangle\in\Bbb F_2[x]/\langle x^3+x+1\rangle$. From now on we remove the lines over the elements and the meaning shall be clear from the context.
We thus have that
$$G:=\text{Aut}\,\left(\Bbb F_8/\Bbb F_2\right)=\{1=Id.\,,\,\sigma\,,\,\sigma^2\}\;,\;\;\sigma x:=x^2$$
So, for example:
$$\begin{align}&\text{Tr.} (x):=Id.x+\sigma x+\sigma^2 x=x+x^2+x^2+x=0\\
&\text{Tr.}(x+1)=x+1+x^2+1+x^2+x+1=1\\\end{align}$$
and etc. You can see in this case the trace is just the sum of the conjugates of each element.
A: You can take advantage of the linearity of trace. To make the notation less cumbersome I denote the coset $x+\langle x^3+x+1\rangle$ by $\alpha$. In that case the elements of the field are
$$
\Bbb{F}_8=\{a_0+a_1\alpha+a_2\alpha^2\mid a_0,a_1,a_2\in\Bbb{F}_2\}.
$$
The minimal polynomial of $\alpha$ is $m(x)=x^3+x+1$. Its other zeros are the conjugates $\alpha^2$ and $\alpha^4=\alpha\cdot\alpha^3=\alpha(\alpha+1)$. Therefore
$$
\begin{aligned}
m(x)&=(x-\alpha)(x-\alpha^2)(x-\alpha^4)=x^3-(\alpha+\alpha^2+\alpha^4)x^2+\cdots\\
&=x^3-tr(\alpha)x^2+\cdots
\end{aligned}
$$
Therefore we can conclude that $tr(\alpha)=0$. This also follows from the earlier calculation (that also appears in Timbuc's answer) giving us that $\alpha^4=\alpha+\alpha^2$.
Because $m(x)$ is also a minimal polynomial of $\alpha^2$, we see that $tr(\alpha^2)=tr(\alpha)$, so also $tr(\alpha^2)=0$. Clearly
$$
tr(1)=1+1^2+1^4=3=1.
$$
Linearity thus gives us that
$$
tr(a_0+a_1\alpha+a_2\alpha^2)=a_0\, tr(1)+a_1\, tr(\alpha)+a_2\, tr(\alpha^2)=a_0.
$$
