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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?

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    Can I ask what the topology is? – 2017-02-12
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    The ring $A$ being connected means that it has no idempotents other than $0$ and $1$. Equivalently it means that its spectrum is connected with respect to the Zariski topology. – 2017-02-14

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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 $a+ b = 1$.

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    I feel silly now, thank you for your clear answer! – 2017-02-17