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 💡