TITLE: Unique factorization domain that is not a Principal ideal domain
QUESTION [13 upvotes]: Let $c$ be an integer, not necessarily positive and not a square. Let $R=\mathbb{Z}[\sqrt{c}]$ 
denote the set of numbers of the form $$a+b\sqrt{c}, a,b \in \mathbb{Z}.$$ 
Then $R$ is a subring of $\mathbb{C}$ under the usual addition and multiplication.
My question is: if $R$ is a UFD (unique factorization domain), does it follow that
it is also a PID (principal ideal domain)?

REPLY [14 votes]: Yes, because such quadratic number rings are easily shown to have dimension at most one (i.e. every nonzero prime ideal is maximal). But $\rm PID\:$s are precisely the $\rm UFD\:$s having dimension $\le 1\:.\ $ Below is a sketch of a proof of this and closely related results.
THEOREM $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ 
$1)\ $ prime ideals are maximal if nonzero
$2)\ $ prime ideals are principal
$3)\ $ maximal ideals are principal
$4)\ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ 
$5)\ $ $\rm D$ is Bezout
$6)\ $ $\rm D$ is a $\rm PID$  
Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)
$1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ 
$2\Rightarrow 3)$ $\ \: $ Clear.
$3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ 
$4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ 
$5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$