Calculate the trace of all elements in $F_8$

Question

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!

Answer 1

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 $$ tr(a_0+a_1\alpha+a_2\alpha^2)=a_0\, tr(1)+a_1\, tr(\alpha)+a_2\, tr(\alpha^2)=a_0. $$

Answer 2

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.