Is $\mathbb{Z}[\sqrt{29}] $ a PID 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?
 A: 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. 
A: Theorem 1: In a PID D, an element is prime iff it is irreducible element in D.  
Define a multiplicative norm : $N(x) : \mathbb Z[\sqrt 29] \to \mathbb Z$ as $N(x) = a^2-29b^2$ where $x= a+b\sqrt 29$.  
Theorem 2: If $x \in \mathbb Z[\sqrt d]$ is unit then $N(x)=1$, and conversely. Consider $d \gt 1$ and a square free integer.   
Theorem 3: If $x \in \mathbb Z[\sqrt d]$ such that $N(x)=p$, where $p$ is a prime, then $x$ is irreducible in $\mathbb Z[\sqrt d]$ .  
Now verify $2 \in \mathbb Z[\sqrt 29]$ is an irreducible element using theorem 2 and the property of multiplicative norm that is,  $N(ab)= N(a)N(b)$.  
But, 2 is not a prime element as $2 \mid (1+\sqrt 29)(1-\sqrt 29)$ but neither $2 \nmid (1+\sqrt 29)$ nor $2 \nmid (1-\sqrt 29)$. So, from theorem 1 you can show that $\mathbb Z[\sqrt 29]$ is not a PID.
