If the roots of a monic irreducible polynomial over ZZ all have modulus at most 1, then the roots are roots of unity

Let p(x) \in \mathbb{Z}[x] be a monic irreducible polynomial and suppose that for each root \zeta of p(x), we have ||\zeta || \leq 1, where || \cdot || denotes complex modulus. Show that the roots of p(x) are roots of unity.

Note that p(x) factors as p(x) = \prod (x - \zeta_i), where \zeta_i ranges over the roots of p. Now the constant coefficient of p(x) is \prod \zeta_i, and thus ||\prod \zeta_i|| = \prod ||\zeta_i|| \leq 1. On the other hand, the constant coefficient is an integer, and so is either 0 or 1. If the constant coefficient is 0, then x divides p(x). So the constant coefficient of p is 1, and we have ||\zeta_i|| = 1 for all i.

By Lemma 11.8, the \zeta_i are all roots of unity.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: