Number Fields

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 $u\prime /\overline{u\prime }$ for all conjugates $u\prime$ 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 $2u^p=0(\mathrm{mod}p)$, so $p$ divides $u$ which is a contradiction since $u$ is a unit. $\blacksquare$