Suppose that $q=2$ and $n=31$.
$C_0=\{0\}$
$C_1=\{1,2,4,8,16\}=C_2=C_4=C_8=C_{16}$
$C_3=\{3,6,12,24,17\}=C_6=C_{12}$
$C_5=\{5,10,20,9,18\}=C_9=C_{10}$
$C_7=\{7,14,28,25,19\}=C_{14}$
$C_{11}=\{11,22,13,26,21\}=C_{13}$
$C_{15}=\{15,30,29,27,23\}$
$ord_{31}(3)=5$, $x^{31}-1$ has roots in $\mathbb{F}_{2^5}$. Let $\alpha$ be the primitive element in $\mathbb{F}_{2^5}$.
Monic polynomial of $C_1$ will be $(x-1)(x-2)(x-4)(x-8)(x-16)=-1024+1984 x-1240 x^2+310 x^3-31 x^4+x^5\equiv30-30x^2+x^5\mod{13}$
But this monic polynomial isn't right, can someone give me a hint or suggestion to get a right a one so that I can construct the generator polynomial? Thanks
If by generator polynomial of $C_1$ you mean the minimal polynomial of $\alpha$ over $\mathbb F_2$, that is, the binary polynomial of least degree that has $\alpha$ as a root, then the polynomial is $$(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)(x-\alpha^{16}).\tag{1}$$ It is also the minimal polynomial of $\alpha^2$ and of $\alpha^4$ and of $\alpha^8$ and of $\alpha^{16}$. The polynomial is irreducible.
In multiplying out the factors, you need use arithmetic in $\mathbb F_{2^5}$ (and not modulo $13$ or $31$ as you have it) but to set up the arithmetic you need to identify which of the $30$ primitive elements of $\mathbb F_{2^5}$ you are choosing to call $\alpha$. The only way to make this identification is to choose one of the six known irreducible binary quintic polynomials and say that $\alpha$ is one of its roots. Of course, once you have chosen, say, $x^5+x^2+1$ as the polynomial, the immediate need of setting up arithmetic to verify that $(1)$ multiplies out to $x^5+x^2+1$ disappears. You will need the arithmetic to figure out the generator polynomials of $C_3$,$C_5$ etc.
So, how do I know that there are 6 quintics and $x^5+x^2+1$ is one of them? I worked it out in this answer.