TITLE: Connectedness in Zariski topology
QUESTION [9 upvotes]: How to show that, for example, the Zariski topology of a cyclic group ring (thus $\mathbb{Z}[\rho]/(\rho^n-1)$) is connected?
Does this still hold for an Abelian group?
Or in general, how do we determine the connected components for such spaces? (Is it by searching its smallest prime ideals? If so, can you give an explicit example?)
Note that it's not a domain, which makes it not necessarily an irreducible space.
Many thanks.
[edit: it seems that I only need to show there's no idempotent element in this group ring, is that right?]

REPLY [6 votes]: Here's the story for arbitrary $n.$ Let $n\in \mathbb{Z}_{n>1}$ and consider the polynomial $f(X) = X^n - 1.$ Then $f$ factors in $\mathbb{Z}[X]$ as the product of cyclotomic polynomials $\displaystyle\prod_{d|n} \Phi_d.$ As $\mathbb{Z}[X]$ is UFD, it follows that the minimal primes of $\mathbb{Z}[X]$ containing $(f)$ are exactly the ideals $(\Phi_d)$ where $d|n.$ Let $d>1$ be a fixed divisor of $n$ and $p$ be a prime dividing $d.$ Write $d = pd_0.$ 
We claim $V(\Phi_d)\cap V(\Phi_{d_0}) \neq \emptyset.$ Assume this is not the case. Then $(\Phi_d,\Phi_{d_0}) = \mathbb{Z}[X],$ so there exist $q_1,q_2 \in \mathbb Z[X]$ such that 
$$q_1(X)\Phi_d(X) + q_2(X)\Phi_{d_0}(X) = 1.$$
Choose a primitive $d$-th of unity $\zeta_d$ in $\overline{\mathbb{Q}}$ and consider $\mathbb{Z}[\zeta_d] = \mathcal{O}_{\mathbb{Q}(\zeta_d)}.$ Then $\zeta_{d_0} := \zeta_{d}^p$ is a primitive $d_0$-th root of unity. Choose a prime $\mathfrak{p}$ of $\mathbb{Z}[\zeta_d]$ dividing $(p)$. Then $\zeta_d \equiv \zeta_{d_0} \mod \mathfrak{p}.$ It follows 
$$1 = q_1(\zeta_d)\Phi_d(\zeta_d) + q_2(\zeta_d)\Phi_{d_0}(\zeta_d) \in \mathfrak{p},$$
a contradiction. We conclude $V(\Phi_d)\cap V(\Phi_{d_0}) \neq \emptyset.$ 
We now consider the spectrum of $R := \mathbb{Z}[X]/f.$ Let $\pi:\mathbb{Z}[X] \rightarrow R$ be the quotient map and $U_1$ be the connected component of $\pi(\Phi_n)$ in $\mathbf{Spec} R.$ Then, as $U_1$ is closed $V(\pi((\Phi_n))) = \pi(V(\Phi_n)) \subset U_1.$ Let $q$ be a prime divisor of $n.$ Then as $V(\Phi_n) \cap V(\phi_{n/q}) \neq \emptyset,$ so too must $V(\pi(\Phi_n)) \cap V(\pi(\phi_{n/q})) \neq \emptyset.$ It follows $\pi(\Phi_{n/p})$ is contained in $U_1.$ Following in this way inductively we obtain that $U_1$ contains $\pi(\Phi_d)$ for each $d|n.$ But these are precisely the minimal prime ideals of $R.$ Hence we conclude,
$$\mathbf{Spec} R \supset U_1 \supset \displaystyle\bigcup_{d|n} V(\pi(\Phi_d))  = \mathbf{Spec} R$$
and $\mathbf{Spec} R$ is connected.