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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.

  • I find it pretty obvious that a primitive $n$-th root of $1$ should be defined as a root of the $n$-th cyclotomic polynomial $\Phi_n$, rather than just as an element that by some happenstance has $n$-th power $1$ without having any earlier power $1$. In that case, the answer is "yes" because the product $\prod\limits_{i=1}^{n-1} \left(1-\zeta_n^i\right) = n$ clearly has the factor $1 - \zeta_n$. – darij grinberg Oct 25 '17 at 23:35
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    (The classical proof of the identity $\prod\limits_{i=1}^{n-1} \left(1-\zeta_n^i\right) = n$, just in case someone hasn't seen it: Start with the polynomial identity $\left(x-1\right) \prod\limits_{i=1}^{n-1} \left(x-\zeta_n^i\right) = \prod\limits_{i=0}^{n-1} \left(x-\zeta_n^i\right) = x^n - 1$, and take derivatives at $x = 1$ on both sides.) – darij grinberg Oct 25 '17 at 23:39
  • I think that if $R$ is local, you also have the answer "yes". Indeed, let $d$ be the order of $\overline{\zeta_n}$ in the multiplicative group of the residue field $R / \mathfrak{m}$. This is finite (since $\zeta_n^n = 1$), and in fact $d \mid n$ (for the same reason). Thus, $d$ is invertible in $R$, and therefore also in $R / \mathfrak{m}$. Hence, since $R / \mathfrak{m}$ is an integral domain, you can use Hurkyl's answer (with $d$ and $R / \mathfrak{m}$ and $\overline{\zeta_n}$ instead of $n$ and $R$ and $\zeta_n$) to show ... – darij grinberg Oct 25 '17 at 23:47
  • ... that $1 - \overline{\zeta_n}$ is invertible in $R / \mathfrak{m}$. Hence, $1 - \zeta_n \notin \mathfrak{m}$, whence $1 - \zeta_n$ is also invertible in $R$. – darij grinberg Oct 25 '17 at 23:48

2 Answers2

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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 $R$)

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    A good catch! ${}$ – Jyrki Lahtonen May 05 '17 at 07:38
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    I expect that the OP's claim is also false when $R$ is a noncommutative as well, maybe, even if $R$ is an integral ring. – Batominovski May 05 '17 at 10:28
  • Nice! That's an excellent point. What if we assume that $R$ is local? (In your counterexample the issue is that there are two connected components, and on one of them $t = 1$, hence on that component $t$ is not a primitive 2nd root of 1) –  May 05 '17 at 17:18
  • Plugging in $x=1$ makes the left hand side $n$ -- what am I missing here? – MT_ Oct 25 '17 at 20:20
  • @MCT: That you can multiply by $n^{-1}$ to get a formula of the form $(1 - \zeta_n) \cdot (\text{something in the ring}) = 1$. –  Oct 25 '17 at 22:48
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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.

Batominovski
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    That's not quite true. In $\mathbb{Q}[t]/(t^2 - 1)$, $t$ is a primitive square root of unity, but $(x-t) \neq 1 + x$. –  May 05 '17 at 07:27
  • Your noncommutative counterexample is actually commutative, since the $\mathbb{K}$-subalgebra of the matrix ring generated by $\Xi_n$ is commutative (and everything is going on in this $\mathbb{K}$-subalgebra). – darij grinberg Oct 25 '17 at 23:41