<s>[INST] Find the inverse in $F = \frac{\Bbb{Z}_{7}[x]}{g(x)}$ of a polynomial $p(x)$ $g(x) = x^{3}+2x^{2}+1$ 
$p(x) = x^{2}+x+2$
I don't know what I did wrong, hope you figured it out:
$g(x) = p(x)(x+1) + 4x + 6$
$p(x) = (4x+6)(2x+6)+1$
$1=p(x)-(2x+6)(4x+6)$
$1=p(x)-(2x+6)[g(x) - p(x)(x+1)]$
$1=p(x)+(5x+1)[g(x) + p(x)(6x+6)]$
$1=(5x+1)g(x) + 2(5x+1)(6x+6)p(x)]$
$1=(5x+1)g(x) + 2(2x^{2}+x+6)p(x)]$
$1=(5x+1)g(x) + (4x^{2}+2x+5)p(x)]$
So: my solution is $(4x^{2}+2x+5)$, but the text book report another result. Any suggestion?
Also: how can I say, without a doubt, that $F$ is a field?
 [/INST] $F$ can be seen as the algebra generated by the matrix $M = \begin{pmatrix}
0 & 0 & -1 \\
1 & 0 & 0 \\
0 & 1 & -2  \\
 \end{pmatrix}$( with entries in $GF(7)\; $), called the companion matrix of the polynomial $g(x) = x^3 + 2x^2 + 1$. This matrix describes the action of the multiplication with $x$ namely $1 \mapsto x$, $x \mapsto x^2$ and $x^2 \mapsto x^3=-1-2x^2$. Then the polynomial $p(x)=x^2+x+2$ is represented by the matrix $N = M^2+M+2M^0 = \begin{pmatrix}
2 & -1 & 1 \\
1 & 2 & -1 \\
1 & -1 & 4  \\
 \end{pmatrix}$. The inverse of $p$ is then represented by the inverse $N^{-1} = \begin{pmatrix}
0 & -2 & 3 \\
1 & 0 & 5 \\
2 & 4 & -1  \\
 \end{pmatrix} = 2M^2 + M $ (note that the coefficients of an element of the algebra in the basis $\{M^0, M, M^2\}$ always can be read from the first column) which corresponds to the polynomial $2x^2+x$.
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