Is $1-\zeta_n$ a unit in every ring where $n$ is invertible? Let $R$ be a commutative ring where $n\ge 2$ is invertible and containing a primitive $n$th root of 1, called $\zeta_n$, satisfying $\zeta_n^n = 1$ and $\zeta_n^k\ne 1$ for any $1\le k\le n$.
Is $1-\zeta_n$ invertible on $R$?
Thanks to Hurkyl's excellent point, one can consider the counterexample $k[t]/(t^2-1)$, where $t$ is a primitive square root of 1, but $t-1$ is a zero divisor, hence not a unit.
In this example, my impression is that you get two connected components of $\text{Spec }k[t]/(t^2-1)$, where on one of them $t = 1$, and on the other $t = -1$. Thus, we have the follow up question:
If $R$ is a local ring, must $1-\zeta_n$ be invertible on $R$?
Also I don't really understand why this was put on hold. The question was apparently clear enough to garner short and yet valuable answers.
 A: Here is a counterexample: in the ring $\mathbb{Q}[t]/(t^2 - 1)$, $t$ is a primitive square root of unity. However, $1-t$ is a zero divisor, and thus not invertible.

The conjecture is true in a domain, however. The powers of $\zeta_n$ are roots of $x^n - 1$, so in a domain you can immediately conclude
$$ \frac{x^n - 1}{x-1} = \prod_{k=1}^{n-1} (x - \zeta_n^k)$$
Plugging in $x=1$ lets you write down an inverse for $1 - \zeta_n$. Explicitly, the left hand side evaluates to $n$ because
$$ \frac{x^n - 1}{x-1} = 1 + x + x^2 + \ldots + x^{n-1}  $$
and so the formula can be rearranged to
$$ (1 - \zeta_n) \cdot \left( n^{-1} \prod_{k=2}^{n-1} (1 - \zeta_n^k) \right) = 1 $$

Another case where the conjecture is true is when $\zeta_n$ is a root of the $n$-th cyclotomic polynomial. In this case, one can construct a ring homomorphism
$$ \mathbb{Z}[\xi_n, n^{-1}] \to R : \xi_n \mapsto \zeta_n $$
where $\mathbb{Z}[\xi_n]$ is the $n$-th ring of cyclotomic integers. Since $1 -
 \xi_n$ is a unit in the cyclotomic integers after inverting $n$, its image $1 - \zeta_n$ must be a unit in $R$.
An important special case where this happens is when $\zeta_n$ is a principal root of unity. (I'm still assuming $n$ is invertible in $R$)
A: Hint:  We have 
$$\prod_{k=1}^{n-1}\,\left(x-\zeta^k_n\right)=\sum_{r=0}^{n-1}\,x^r\,.$$
EDIT:  See Hurkyl's comment below.

A Noncommutative Counterexample 
A positive integer $n>1$ is fixed, and a field $\mathbb{K}$ whose characteristic does not divide $n$ is given.  Let $R$ be the ring of $n$-by-$n$ matrix over $\mathbb{K}$ with the identity $I_n$.  Then, the $n$-by-$n$ permutation matrix $$\Xi_n:=\begin{bmatrix}0&1&0&\cdots&0&0\\0&0&1&\cdots&0&0\\0&0&0&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&0&1\\1&0&0&\cdots&0&0\end{bmatrix}$$ is a primitive $n$-th root of unity.  However, $I_n-\Xi_n$ is not invertible, having the $\mathbb{K}$-span of the $n$-by-$1$ column vector $$\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix}$$ as the nullspace.  I'm curious whether there is a counterexample with a noncommutative integral ring.
