Prove $\{x\in R | x^n-1=0\}=\{1,\omega,\dots,\omega^{n-1}\}.$ $R$ is any commutative ring with $1$ (a famous algorithm book by Aho, Hopcroft, Ullman)

Question

I am reading "The Design and Analysis of Computer Algorithms" by Aho, Hopcroft, Ullman.

Without a proof, the authors wrote the following proposition held.

Let $R$ be any commutative ring with $1$.
Let $\omega$ be an element of $R$ which satisfies the following three conditions:

1. $\omega\ne 1.$
2. $\omega^n=1.$
3. $1+\omega^p+(\omega^p)^2+\dots+(\omega^p)^{n-1}=0$ for all $p\in\{1,\dots,n-1\}.$

Then, $\{x\in R | x^n-1=0\}=\{1,\omega,\dots,\omega^{n-1}\}.$

I can prove the above proposition holds if $R$ is an integral domain, but I cannot prove that the above proposition holds for all commutative rings with $1$.

Answer 1

Indeed, the claim is false. A counterexample is $R=\Bbb Z/12\Bbb Z$ and $\omega=-1$ and $n=2$. The three conditions are obviously true, yet $\{x\in R\colon x^2-1=0\} = \{1, 5, -5, -1\}$. (This remains a counterexample when $12$ is replaced by any integer with at least two distinct prime factors that is either odd or a multiple of $4$.)