Given $\,f(t)=t^3+t+1,\,$ let $\,\mathbb{F}_{2^3}=\frac{\mathbb{F}_2\left[t\right]}{\langle f\,\rangle}\,$ be a finite field and $\,\xi=t\,\left(\textrm{mod}\,f\right).$ Let also $\,\mathcal{C}\,$ be a cyclic code where $\,g(x)=\left(x-\xi\right)\left(x-\xi^2\right)\left(x-\xi^3\right)\left(x-\xi^4\right)\,$ is his generator polynomial.
Question: Find a generator matrix of $\,\mathcal{C}\,$ in the basis $\,B=\left\{1,\xi,\xi^2\right\}\,$ of $\,\mathbb{F}_{2^3}\,$ as an $\,\mathbb{F}_{2}$– vector space.
The fisrt thing they ask is to prove that $\,\xi\,$ is a $\,7^{\,th}\,$ primitive root of the unity (quite easy). Then, prove that the length of $\,\mathcal{C}\,$ is $\,n=7$.
I found out that $\,\mathcal{C}\,$ is a Reed-Solomon code, a type of BCH code whose length is always $\,n=q-1\,$ over a finite field $\,\mathbb{F}_q,\,$ so this question is also easy to answer.
However, I do not know how to proceed to find the generator matrix of $\,\mathcal{C}.$ I tried to expand the generator polynomial
$$g(x)=g_0+g_1x+\dots+g_4x^4$$
and define the generator matrix as
$$G=\begin{pmatrix}g_0&g_1&g_2&g_3&g_4 & 0 & 0\\
0 & g_0&g_1&g_2&g_3&g_4 & 0\\
0 & 0 & g_0&g_1&g_2&g_3&g_4\end{pmatrix}$$
in the basis $\,B=\left\{1,\xi,\xi^2\right\}\,$ but I guess I failed.
Somebody please, help me! Thanks in advance.
Best Answer
Lots of comments. We have $$g = (x-\xi)(x-\xi^2)(x-\xi^3)(x-\xi^4).$$ If $\xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $\xi,\xi^2,(\xi^2)^2=\xi^4$. Thus $$g = (x^3+x+1)(x+\xi^3) = x^4+x^2+x+\xi^3x^3+\xi^3 x +\xi^3\\ = x^4 + \xi^3 x^3 +x^2+(\xi^3+1)x+\xi^3.$$ Since $\xi$ is a zero of $f$, $\xi^3+\xi+1=0$ and so $\xi=\xi^3+1$ Hence,
$$g= = x^4 + \xi^3 x^3 +x^2+\xi x+\xi^3.$$