# Minimum absolute values of algebraic integers in cyclotomic fields

Let $\omega$ be a primitive $p^{th}$ root of unity with $p$-odd prime.

Consider $\mathbb{Z}[\omega]$, the ring of integers in cyclotomic field $\mathbb{Q}(\omega)$. I wanted to know how small can be the elements of $\mathbb{Z}[\omega]$, in the sense -

Is the set $$\{|\alpha|: \alpha\in\mathbb{Z}[\omega], \alpha\neq 0\}$$ bounded below?

• I am considering absolute values as considering the element to be a complex number. Dec 6, 2016 at 4:18

## 2 Answers

It can be as small as you want. Take $p=5$, so $\omega^2+\omega^3=-\frac 12(1+\sqrt 5)$. As this is irrational, by the equidistribution theorem we can find integers $a,b$ so that $a-b\frac 12(1+\sqrt 5)$ is as small as we want.

If you want something less beefy than the equidistribution theorem, just take $p$ to be whatever you like. You know that $\sqrt{p^*}\in\Bbb Z[\omega]$ where $p^*=(-1)^{p-1/2}p$, and since $p^*\equiv 1\mod 4$ we have ${2j+1+\sqrt{p^*}\over 2}\in\Bbb Z[\omega]$ is an algebraic integer for all $j\in\Bbb Z$. If $p\equiv 1\mod 4$ then let $n<\sqrt{p^*}=\sqrt{p}<n+1$. There are two cases:

$$\alpha = \begin{cases} {n-\sqrt{p^*}\over 2} & n = 2k+1 \\ {n+1-\sqrt{p^*}\over 2} & n= 2k \end{cases}$$

both of which give a number, $\alpha$, with absolute value $<1$, in particular its powers converge to $0$, which gives an infinite family of rings with small absolute values.