# Connectedness in Zariski topology

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?]

For an Noetherian ring $R$, $X=\mathrm{spec}(R)$ is connected if and only if the only idempotent elements of $R$ are $0,1$, or equivalently, $R$ is not of the form $R_1\times R_2$ for two subring $R_1,R_2$.

For if $X$ is disconneted, it can be written as the disjoint union of two proper closed subspaces defined by ideal $I_1, I_2$ respectively. Since $V(I_1)\cup V(I_2)= X$, We must have $I_1I_2$ is an ideal contained in the nilradical of $R$. Replacing $I_i$ by a power of $I_i$, we may assume that $I_1I_2=0$ (note that we use the Noetherian assumption here). On the other hand, $V(I_1)\cap V(I_2)=\emptyset$, so we get $I_1+I_2=R$. In particular, we get $I_1=I_1(I_1+I_2)=I_1^2$. Since $I_1$ is finitely generated, it follows from Cayley-Hamilton that one elements $e_1\in I_1$ acts as identity on $I_1$. In particular $e_1\cdot e_1=e_1$. We take $e_2=1-e_1$. Let $R_1=Re_1, R_2=Re_2$, then $R$ can be written as the direct product of the two subrings.

But in general, I am afraid that I don't know ways to find idempotents except the add-hoc methods. For example, in your case, if $n=p^r$ is a prime power, then we know the fiber over $p$ is connectd (because it is spec of $\mathbb{F}_p[x]/((x-1)^q)$). If $X$ is disconnected, then we must have $e_1 \equiv 0 \mod p$ and $e_2 \equiv 1 \mod p$. But $e_1^2=e_1$, so $e_1 \equiv 0 \mod p^m$ for all $m>0$, which leads to a contradiction. But I am not sure how one generalize this to the case $n$ is arbitrary.

Edit: I see, you already got the idempotent part.

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.

• Take $d=6$ and $p=2$, then $\zeta_6-\zeta_6^2=\zeta_6(1-\zeta_6)$ is a unit. But you are still right in the sense that we will have $$0=\Phi_{d_0}(\zeta_{d_0})=\Phi_{d_0}(\zeta_{d}^p)\equiv \Phi_{d_0}(\zeta_{d})^p \pmod{\mathfrak{p}},$$ hence $\Phi_{d_0}(\zeta_{d}) \equiv 0 \pmod{\mathfrak{p}}$. Very nice answer, by the way. – Jiangwei Xue Jul 2 '11 at 17:01