I would like to show that for an cyclic code $C\subset F_q[x]/(x^n -1)$ with check polynomial $p(x)=p_0+p_1x+...+p_k x^k$ there exists this check matrix:
$$ \begin{pmatrix} p_k & \cdots & p_1 & p_0 & 0 & \cdots & 0 \\ 0 & p_k & \cdots & p_1 & p_0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots &\ddots & \ddots & 0 \\ 0 & \cdots & 0 & p_k & \cdots & p_1 & p_0 \\ \end{pmatrix} $$
Here's my approach: Lets call that Matrix $P$. $P$ is a check matrix of $C$, when $v\in C \Leftrightarrow P\cdot v^T=0$.
Let $v\in F_q^n, v=(v_0,...,v_{n-1})\\ H\cdot v^T=0 \Leftrightarrow 0=\sum_{i=0}^{k}p_{k-i}\cdot v_{i+h} \forall h \in \{0,...,n-k-1\}\\ \Leftrightarrow 0=\sum_{j+l=k+h}p_j\cdot v_l \forall h\\ \Leftrightarrow \text{all coefficients of }x^k,...,x^{n-1} \in p(x)\cdot v(x) \text{ are zero mod }x^n-1 $
Here im stuck. I want to follow, that $p(x)\cdot v(x) \equiv 0 \text{ mod } x^n-1$. But i don't know how to show that.
Because then i can follow that $P$ is check matrix if and only if $v \in C \Leftrightarrow P\cdot v^T =0 \Leftrightarrow p(x)\cdot v(x)\equiv 0$.
If the check polynomial for you is the polynomial p(x) whose product with every c(x) in the code is zero, then the statement is wrong. The dual is generated by the reciprocal of h(x). For instance the scalar product of p and c as vectors is the constant term of $p^*(x)c(x).$