Every nonzero ideal in a ring of algebraic integers contains infinitely many rational integers

Let \mathcal{O} be the ring of integers in a fixed finite extension K = \mathbb{Q}(\theta), and let A be a nonzero ideal in \mathcal{O}. Prove that A contains infinitely many rational integers.

We begin with a lemma.

Lemma: Let A \subseteq \mathcal{O} be an ideal. If \alpha \in A, then N(\alpha) \in A, where N denotes the field norm. Proof: Note that N(\alpha) \in K since N(\alpha) is a rational integer. So N(\alpha)/\alpha \in K. Moreover, by definition, N(\alpha)/\alpha is an algebraic integer, being the product of the (certainly integral) conjugates of \alpha for K. So N(\alpha)/\alpha \in \mathcal{O}. Since A is an ideal, \alpha(N(\alpha)/\alpha) = N(\alpha) \in A. \square

Now if A \neq 0, then there exists a nonzero element \alpha \in A. We have N(\alpha) \in A nonzero, and thus kN(\alpha) \in A for all k \in \mathbb{Z}. In particular, A contains all of the \mathbb{Z}-multiples of N(\alpha).

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: