TITLE: Is $\mathbb{Z}[\sqrt{29}] $ a PID
QUESTION [6 upvotes]: Question as in title. I think that unique factorization fails, perhaps via either $ (\sqrt{29} - 1)(\sqrt{29} + 1) = 2^2 \cdot 7 $ or $ (\sqrt{29} - 5)(\sqrt{29} + 5) = 2^2 $, but I have trouble proving either of these two claims. How to solve this problem?

REPLY [3 votes]: Any UFD is normal, i.e. it equals its integral closure in its field of fraction. Here the integral closure of $ \mathbb{Z}[\sqrt{29}] $ in $ F $ contains the integral closure of $ \mathbb{Z}$ in $ F $, which is strictly larger than $ \mathbb{Z}[\sqrt{29}] $. Hence this ring is not normal, so it is not a UFD and therefore not a PID.