Prime ideals of $F_q[C_m]$ My main question is that what the prime ideals of group ring $F_q[C_m]$ are where $F_q$ is a finite field with $q$ elements and $C_m$ is a cyclic group of order m. To do this, I was thinking that how the ideals of $F_q[C_m]$ are. I could find just one non trivial ideal for arbitrary group ring $R[G]$ which is the set of all elements of the form $\Sigma_{g\in G}ag$ where $a\in R$. Also, I tried some easy cases for small q, m which I understood there are a lot of computations, to find at least an ideal in them. 
 A: In the case of our group being $C_m$, we can recognise this ring as something more familiar, letting us see its structure (and thus its ideals). We have a surjective map $\mathbf{F}_q[x]\rightarrow \mathbf{F}_q[C_m]$ given by mapping $x$ to a generator of $C_m$, and we see that $x^m-1$ is the kernel of this map. So by the first isomorphism theorem, it factors through this quotient $\mathbf{F}_q[x]/(x^m-1)\rightarrow \mathbf{F}_q[C_m]$, and by counting dimensions, we see that this is a ring isomorphism, since it's surjective.
So now $\mathbf{F}_q[x]$ is a principal ideal domain, so we can factorise $x^m-1=\prod_i f_i(x)^{e_i}$, where each $f_i$ is an irreducible polynomial. Since the ideals generated by these ideals are coprime, the chinese remainder theorem lets us identify the quotient $\mathbf{F}_q[x]/(x^m-1)$ as $\prod_i \mathbf{F}_q[x]/f_i(x)^{e_i}$. So we only need to understand the ideals in the ring $\mathbf{F}_q[x]/f(x)^e$, for $f(x)$ an irreducible polynomial. This ring has a unique prime ideal, given by $(f(x)^{e-1})$, and one can show that all ideals are of this form $((f(x)^k)$ generated by a power of $f(x)$.
So putting this together, we see that the number of prime ideals of $\mathbf{F}_q[C_m]$ is equal to the number of irreducible factors of $x^m-1$ in $\mathbf{F}_q$, and we will see nilpotent elements in $\mathbf{F}_q[C_m]$ if and only if we have irreducible factors with multiplicity.
This case is somewhat deceptive, in general, finding the prime ideals of $k[G]$ is much harder, when $G$ is noncommutative, this requires the representation theory of finite groups.
