# Definition of $\mathbb{Z}[\omega]$ where $\omega$ is a primitive root of unity

What does $\mathbb{Z}[\omega]$ usually mean when $\omega$ is a primitive root of unity?

-
There is a nice explanation available here: artofproblemsolving.com/blog/14170 . I hope that helps :) – WWright Dec 2 '10 at 4:37

## 2 Answers

$\mathbb{Z}[\omega]$ is the ring generated by $\mathbb{Z}$ and $\omega$ (inside, say $\mathbb{C}$). You can think of this as a ring whose elements are polynomials in $\omega$ with coefficients in $\mathbb{Z}$. Since, $\omega^n=1$ for the appropriate $n$, you need to only consider polynomials of degree less than $n$. Then the usual addition and multiplication of polynomials give you the ring structure.

-
Thanks. That is what I was thinking it meant. – Digital Gal Dec 2 '10 at 4:44
Careful; this explanation sounds like it's implying that $\mathbb{Z}[\omega] \cong \mathbb{Z}[x]/(x^n - 1)$, but this is wrong; the latter has zero divisors while $\mathbb{Z}[\omega]$ doesn't. In fact the minimal polynomial of a primitive $n^{th}$ root of unity is given by the cyclotomic polynomial $\Phi_n(x)$ (en.wikipedia.org/wiki/Cyclotomic_polynomial). – Qiaochu Yuan Feb 18 '12 at 22:42

When $\omega$ is any algebraic element of $\mathbb{C}$ satisfying a monic polynomial with integral coefficients, $\mathbb{Z}[\omega]$ denotes the ring $\{a_0+a_1\omega+\ldots+a_{n-1}\omega^{n-1}: a_i\in \mathbb{Z}, n=\text{deg }\omega\}$, where the degree of $\omega$ is the degree of the minimal polynomial of $\omega$ over $\mathbb{Q}$. When $\omega$ is a primitive $n$-th root of unity, this ring happens to be the ring of integers of the cyclotomic field $\mathbb{Q}(\mu_n)$, but for general algebraic $\omega$, $\mathbb{Z}[\omega]$ is a proper finite index subring of the ring of integers of $\mathbb{Q}[\omega]$.

-
rtel: Your first $\mathbb{Z}$ should be $\mathbb{Z}[\omega]$; I think you mean the expression to go only up to $\omega^{n-1}$; and "monic" (not "monimal"). But if the monic polynomial does not have coefficients in $\mathbb{Z}$, then your set is not a ring (it's not closed under multiplication). – Arturo Magidin Dec 2 '10 at 4:42
Thanks a lot, Arturo! I had caught some of these mistakes within the 5 minute grace period, but obviously not all :-) – Alex B. Dec 2 '10 at 4:45