A bound on the number of irreducible factors of an element in an algebraic integer ring

Let \alpha be an algebraic integer in an algebraic number field K. Let N denote the norm over K. Prove that the number of factors in any irreducible factorization of \alpha over K is at most \log_2|N(\alpha)|.


Note that \log_2|N(\alpha)| is some real number between t and t+1, where 2^t \leq |N(\alpha)| < 2^{t+1}. No natural number less than 2^{t+1} can have t+1 prime factors (including multiplicity), and since N is multiplicative, the number of irreducible factors of \alpha is bounded above by the number of prime factors of N(\alpha). So the number of irreducible factors of \alpha in K is at most \log_2|N(\alpha)|.

Advertisements
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: