# Algebraic Integers in Cyclotomic Field

Let $\zeta^n=1$, assume that $\displaystyle\alpha=\frac{\sum_{i=1}^m\zeta^{n_i}}{m}$ is an algebraic integer. Show that either $\alpha=\zeta^{n_i}$ for each $i$ or $\alpha=0$.

I am confused with the notation $\zeta^n$. $\zeta$ should not be primitive, but how can I tackle with the powers $n_i$?

Any help is appreciated!

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One way of stating this is: If an algebraic integer $\alpha$ is the average of some sequence of (possibly non-primitive) $n$th roots of $1$, then $\alpha$ is either an $n$th root of $1$ or zero. – Thomas Andrews Mar 27 '12 at 13:36

## 1 Answer

Note by simple geometry, if the $n_i$ is different, then $\alpha$ is strictly inside the unit circle in the complex plane.

But so are all its $\phi(n)$ conjugates in the cyclotomic field, because the conjugates are all also averages of $m$ $n$th roots of unity, not all equal.

So the norm of $\alpha$ in the cyclotomic field has absolute value less than $1$. For $\alpha$ to be an algebraic integer, its norm must be an integer, and hence its norm must be zero. Therefore, $\alpha=0$.

So the only way for $\alpha$ to have non-zero norm is if all the $n_i$ are equal, and then $\alpha=\zeta^{n_i}$.

So the two things you need to know are:

1. A specific "strong" convexity condition on the closed unit disk - the average of points on the boundary are strictly inside the unit disk unless all the points are equal.
2. The automorphisms of the cyclotomic field send averages of $n$-th roots of unity to averages of $n$-roots of unity.
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Thomas, that's a great answer, a thousand thanks! – Qiang Zhang Mar 28 '12 at 1:01