I've got some troubles solving an (easy ?) exercise:
Let n=26, q=3. Now I'd like to find the cyclotomic cosets modulo n over $F_q$.
The cyclotomic cosets are defined as $C_s = (s, ns, n^2s, ..., n^{q-1}s)$, right?
So in my example, the cyclotomics would be:
$C_0 = \{0\}$ $C_1 = \{1, 26, 676\}$
Is this correct until now? I'm really not sure about the 676... And often (in literature), there is no $C_2$, $C_4$ etc. --> But why?
Thanks for your help!
From the discussion in the comments, it turns out that the question concerns the cyclotomic cosets of $3$ modulo $26$. So, first we look at the powers of $3$, modulo $26$. Since $3^3=27\equiv1\bmod{26}$, the powers of $3$ modulo $26$ are just $\{\,1,3,9\,\}$. So this is $C_1$ (and it's also $C_3$, and $C_9$). Then
$C_2=\{\,2,6,18\,\}$, $C_4=\{\,4,12,10\,\}$, $C_5=\{\,5,15,19\,\}$, $C_7=\{\,7,21,11\,\}$, $C_8=\{\,8,24,20\,\}$, $C_{13}=\{\,13\,\}$, $C_{14}=\{\,14,16,22\,\}$, $C_{17}=\{\,17,25,23\,\}$
and I suppose you need $C_0=\{\,0\,\}$.