Unique factorization domain that is not a Principal ideal domain

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)?

Solution 1:

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
$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$

Solution 2:

The answer is yes. The argument is as follows: if $R$ is a UFD, then it is necessarily integrally closed in its fraction field $K = \mathbb Q(\sqrt{c})$, and thus is equal to the full ring of algebraic integers in $K$. A general fact about such full rings of algebraic integers is that if they are UFDs then they are PIDs, the reason being that in these rings, one always has the unique factorization of non-zero ideals into prime ideals, and it is not hard to see that the UFD property forces prime ideals (and hence any product of prime ideals) to be principal.

Solution 3:

Yes. If it is a UFD, then it is integrally closed, hence it is a Dedekind domain because it is of dimension one (being contained in the integral closure of $\mathbb{Z}$ in some finite extension of $\mathbb{Q}$). A Dedekind domain is a UFD iff it is a PID: indeed, this is equivalent to every non-zero prime being principal. (A noetherian domain is a UFD iff every height one prime is principal. So if a Dedekind domain is a UFD, then all its primes are principal, so by factorization of ideals, every ideal is principal.)

A simple example of a UFD that is not a PID is the polynomial ring $\mathbb{C}[x,y]$.