## 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.∎

Ben Lynn blynn@cs.stanford.edu 💡