Remark

The discriminant $D$ of $\Phi$ with respect to the basis $(x_j)_{(j=0,\ldots,2r)}$ is equal to $2^{2r+1} Q(x_0) \ldots Q(x_{2r})$. Consequently $Z$ is generated by $1$ and by the odd element $z' = 2^{r+1} z$ such that ${z'}^2 = (-1)^r 2D$. The algebra $C(Q)$ is therefore simple if and only if $2(-1)^r D$ is not a square in $A$.