# 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)?

Yes, because (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 which have dimension $$\le 1.\,$$ Below is a sketch of a proof of this and closely related results.

Theorem $$\rm\ \ TFAE\$$ for a $$\rm UFD\ D$$

$$(1)\ \$$ prime ideals are maximal if nonzero,  i.e. $$\rm\ dim\,\ D \le 1$$
$$(2)\ \$$ prime ideals are principal
$$(3)\ \$$ maximal ideals are principal
$$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1,\,$$ i.e.  coprime $$\Rightarrow$$ comaximal
$$(5)\ \$$ $$\rm D$$ is Bezout, i.e. all ideals $$\,\rm (a,b)\,$$ are principal.
$$(6)\ \$$ $$\rm D$$ is a $$\rm PID$$

Proof $$\$$ (sketch of $$\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\$$ where $$\rm\,p_i,\,P\,$$ denote primes $$\neq 0$$

$$(1\Rightarrow 2)$$ $$\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$$ some $$\rm\,p_j\in P\,$$ so $$\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$$ by dim $$\le1$$
$$(2\Rightarrow 3)$$ $$\$$ max ideals are prime, so principal by $$(2)$$
$$(3\Rightarrow 4)$$ $$\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p)$$ for all max $$\rm\,(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)$$ $$\$$ Ideals $$\neq 0\,$$ in Bezout UFDs are generated by an elt with least #prime factors
$$(6\Rightarrow 1)$$ $$\ \ \rm (d) \supsetneq (p)$$ properly $$\rm\Rightarrow\,d\mid p\,$$ properly $$\rm\,\Rightarrow\,d\,$$ unit $$\,\rm\Rightarrow\,(d)=(1),\,$$ so $$\rm\,(p)\,$$ is max

Remark  Examples of non-PID UFDs are easy in polynomial rings: if $$D$$ is a non-field domain then it has a nonzero nonunit $$d$$ so by here the ideal $$(d,x)$$ is not principal.

• I am not able to understand your 4th condition. I am confused by the notations. What exactly do you want to say.
– anonymous
Commented Jan 20, 2011 at 9:51
• @Chandru1: $\rm\ (a,b) = 1$ means the ideal generated by $\rm\ a,\:b\$ is the whole ring, i.e. $\rm\ a\ d + b\ c = 1\$ for some $\rm\ c,\: d\ \in D\:.\$ $\rm\ \gcd(a,b) = 1\$ means $\rm\ d\ |\ a,\:b\ \Rightarrow\ d\:|\:1\:.$ Commented Jan 20, 2011 at 14:15

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.

• Dear Matt E Maybe I could impose for some pedagogical advice. I'm a 68 year old self-studier, no math education, starting two years ago. I love algebra and what little I know of algebraic number theory. I have often seen your kind and highly informative remarks. E.g., here I am familiar with all the terms used, but am not remotely capable of putting them together as such. I would be privileged if you gave me some advice as to a book/study regime to get some control of the material. I apologize if this is not an appropriate question or venue in which to ask. Thanks very much. Regards
– user12802
Commented Feb 10, 2013 at 15:51
• Dear Matt E Please forgive me for you adding to this imposition, but maybe I could also mention I post I wrote to a question that is most relevant. This is such a big part of my life. meta.math.stackexchange.com/questions/2606/…
– user12802
Commented Feb 11, 2013 at 18:30
• @Andrew: Dear Andrew, Thanks for the kind words. I posted some comments under your meta post which might be of some help. It's normal that the arguments you can work out for yourself lag a long way behind the arguments of others that you can read, but I think working through exercises in a good undergrad algebra book would be a way to improve. As I wrote in my other comments, Dummit and Foote has many good exercises (as far as I remember), especially on Galois theory and related topics, which are good background for the kind of number theory you seem to be interested in. Best wishes, Commented Feb 12, 2013 at 6:26

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 $\mathbb{C}[x,y]$.