Prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element I need to prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element, but I'm not really sure how to begin.
I know that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ consists of all formal sums $w_{1}[z]_{1} + w_{2}[z]_{2} + \cdots + w_{k}[z]_{k}$ where the $w_{i}$ are integers, and the $[z]_{i}$ are elements of $\mathbb{Z}_{n}$. Now, my first question is, should these $[z]_{i}$ actually just be $\{1, 2, \cdots \, n\}$? I think being able to properly visualize what elements of this group ring look like might be rather helpful in figuring out how to do this problem.
Next, I know that $\mathbb{Z}_{n}$ itself can be generated by the single element $[1]_{n}$ and $\mathbb{Z}$ by $1$. How to extend this to a group ring is escaping me, however. 
Any kind of push in the right direction would be welcome at this point by someone patient and willing to answer lots and lots of follow-up questions on my journey to understanding this completely.
Thank you ahead of time!
 A: Let $C_n$ denote the cyclic group of order $n$, with generator $a$. The group ring $\mathbb{Z}[C_n]$ is precisely the ring
$$R := \mathbb{Z}[x]/(x^n-1)$$
You should think of this presentation as saying: It is the ring which contains $\mathbb{Z}$ and contains "$a$", with the property that $a^n = 1$, ie $a^n-1=0$. This gives rise to the relation $x^n-1$. It's easy to check that both $R$ and $\mathbb{Z}[C_n]$ as $\mathbb{Z}$-modules are free of the same rank, so it's reasonable to think that they're isomorphic.
To prove that they are isomorphic, it suffices to define a ring homomorphism from one to the other, and to check that it is an isomorphism. The easiest way to do this would be to define a homomorphism from $\mathbb{Z}[x]\rightarrow \mathbb{Z}[C_n]$ which sends $x\mapsto a$. It's a homomorphism by the universal property of polynomial rings. Then it suffices to check that the kernel is precisely the ideal $x^n-1$. I haven't done this, but I think it should be reasonably straightforward.
