Principal ideals generated by irreducible elements need not be prime

Suppose \pi is an irreducible element in an algebraic integer ring \mathcal{O}. Must (\pi) be prime as an ideal in \mathcal{O}?

Recall that 1+\sqrt{-5} is irreducible in \mathcal{O}_K, where K = \mathbb{Q}(\sqrt{-5}), since no element in this ring has norm 3. Moreover, 1 - \sqrt{-5} \notin (1+\sqrt{-5}), as we show. If to the contrary we have (a+b\sqrt{-5})(1+\sqrt{-5}) = 1 - \sqrt{-5}, then comparing coefficients we have a-5b = 1 and a+b = -1, so that 3b = -2 for some b \in \mathbb{Z}, a contradiction. Using this previous exercise, we have (1+\sqrt{-5}) \subsetneq (1+\sqrt{-5}, 1-\sqrt{-5}) \subsetneq \mathcal{O}. In particular, 1+\sqrt{-5} is irreducible, but (1+\sqrt{-5}) is not maximal and thus not prime.

So it need not be the case that (\pi) is prime if \pi is irreducible.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

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

You are commenting using your 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: