Q:

Given an order and a prime $p$, is its subring generated by $p$-th roots of unity always connected? 

As an exercise I've been asked to prove a statement I believe to be false:

Let $A$ be an order and let $A^{\times}_p$ its $p$-th roots of unity, where $p$ is any prime number.  Show that the subring of $A$ generated by $A^{\times}_p$ is connected. 

Here $A$ being an order means that $A$ is a commutative ring of which the additive group is isomorphic to $\Bbb{Z}^n$ for some natural number $n$, and 
$$A^{\times}_p=\{a\in A:\ a^p=1\}.$$
Having no clue how to prove this I thought I'd try an example; the ring $A=\Bbb{Z}\times\Bbb{Z}$, with $p=2$. Then
$$A^{\times}_2=\{(1,1),(1,-1),(-1,1),(-1,-1)\},$$
but then the subring of $A$ generated by $A^{\times}_2$ contains both
$$(1,1)+(1,-1)=(2,0)\qquad\text{ and }\qquad(1,1)+(-1,1)=(0,2),$$
where $(2,0)\cdot(0,2)=(0,0)$, meaning that this subring isn't connected. 
It seems odd to me that such a simple counterexample should exist.  Is my reasoning correct? 

A:

Your element are not idempotent, and actually your subring is connected.   It is the ring $R := \{(m,n) \in \mathbb Z\times \mathbb Z \, | \, m \equiv n \bmod 2\},$ and if you compute $R/2R$, it is the ring $\mathbb F_2[\epsilon]/(\epsilon^2)$.  From this, it is easy to see that $R$ contains no non-trivial idempotents. 
P.S.  You didn't find orthogonal idempotents, but rather you found elements $a$ and $b$ such that $ab = 0$ and $a+b$ is a non-zero divisor.   This shows that Spec $A$ is not irreducible.   But to get a disconnection of Spec $A$, you need the stronger condition that $a+ b = 1$.

