Write down the $8$ elements of $ F_2[x]/(x^3 + x + 1) $ in terms of α Consider the field $F_2[α] = F_2[x]/(x^3 + x + 1)$, where $α$ is the image of $x$ in $F_2[x]/(x^3 + x + 1)$. Write down the $8$ elements of this field in terms of $α$.
I have no clue how to start this question. Thanks!
 A: The field $F_2$ has two elements, $0$ and $1$.  
Elements of $F_2[\alpha]$ can be expressed as $c_0 + c_1\alpha + c_2\alpha^2,$ with $c_0, c_1,$ and $ c_2\in F_2$.  
Higher powers of $\alpha$ are not needed because $\alpha^3= \alpha+1$ 
(by construction, $\alpha^3+\alpha+1= 0$ so $\alpha^3= -\alpha-1$, and $-1=1$ in $F_2,$ so $\alpha^3= \alpha+1$).  
There are $2$ possibilities for each of $c_0, c_1,$ and $c_2$, so $8=2^3$ possibilities altogether: 
$0, 1, \alpha, 1+\alpha, \alpha^2, 1+\alpha^2, \alpha+\alpha^2,$ and $1+\alpha+\alpha^2$.
A: This is a form of graphical representation complementing the solution by @J. W. Tanner : here are (Fig. 1) the addition table and (Fig. 2) the multiplication table of this field.

Fig. 1 : Addition table. Please note the diagonal of zeros (deep blue) due to the fact that $w+w=0$, whatever $w$... remark as well the partition into blocks $4 \times 4$ : why that ?

Fig. 2 : Multiplication table. Please note the natural bordering by zeros......
How did I obtain them ? Without entering into the details, I will just say that instead of working with polynomials having their coefficients in $\{0,1\}$, I have been using a $3 \times 3$ matrix $M$ such that $M^3+M+1=0 \ \color{'red'}{modulo 2}$. It is easy to find such a matrix (the so-called companion matrix of polynomial $x^3+x+1$). Here it is :
$$M=\begin{pmatrix}0&1&1\\1&0&0\\0&1&0\end{pmatrix}$$
The elements of the field are thus all the $aI_3+bM+cM^2$, $a,b,c \in \{0,1\}$.
