## Kummer’s Lemma

**Lemma:** Let $\omega = e^{2 \pi i/p}$ for some odd prime $p$.
If $u \in \mathbb{Z}[\omega]$ is a unit, then
$u / \bar{u}$ is a power of $\omega$.

[This simple-sounding lemma is more involved than it first appears. At first, I thought it was obvious that any unit is a power of $\omega$, but this is of course obviously false: $-\omega$ is not a power of $\omega$.]

**Proof:** Let $f\in\mathbb{Q}[x]$ be a monic polynomial.
Suppose all the roots of $f$ have absolute value 1. Then the sum of the
roots taking them $r$ at a time is bounded by
$(_r^n)$
by the triangle inequality. Thus the coefficient of $x^r$ is bounded by
this $(_r^n)$, hence for any fixed $n$
there are only finitely many algebraic integers $\alpha$ such that all
conjugates have absolute value 1 because there are only finitely many
polynomials in $\mathbb{Z}[x]$ with given bounded coefficients.

Then consider the powers of $\alpha$. They are all algebraic integers of degree at most $n$, and furthermore all their conjugates also have absolute value 1 since the Galois actions map powers of $\alpha$ to powers of its conjugates. Thus the powers of $\alpha$ are restricted to a finite set. This means $\alpha$ is a root of unity.

Now consider the conjugates of $u/\bar{u}$, that is $u'/\bar{u'}$ for all conjugates $u'$ of $u$. Since complex conjugation is one of the Galois actions they are all algebraic integers with absolute value 1, thus they are roots of unity. Hence $u/\bar{u} = \pm \omega^k$ for some $k$.

Suppose $u/\bar{u} = {-\omega^k}$. Then $u^p = {-\bar{u}^p}$. But modulo $p$, we have $u^p = \bar{u}^p$. Thus $2 u^p = 0 (mod p)$, so $p$ divides $u$ which is a contradiction since $u$ is a unit.∎