I was looking for strategies to prove when rings are going or not to be UFD's. I really only know that if I manage to prove that there is an element on the fraction field $K$ of my ring $R$ that is not integral (meaning that it does not solve a monic polynomial in $R$) then the ring won't be integrally closed and hence not a UFD.

Are there other strategies to do so? (Like for instance to show something is a domain it is sometimes easy to show that it is a ring quotient out by a prime ideal).


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    $\begingroup$ You haven't understood the definition of "integrally closed". Read it again. Also, I don't understand what do you mean with your sentence about checking that something is a domain. $R/I$ is a domain iff $I$ is a prime ideal of $R$, this has absolutely nothing to do with $R$ being a domain or not. Finally, for your question: Have you read en.wikipedia.org/wiki/Unique_factorization_domain ? $\endgroup$ – Martin Brandenburg Sep 9 '13 at 23:16
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    $\begingroup$ Sorry the thing about the domain was a typo. I know that UFD implies Integrally closed. Hence, if we manage to show that there is an element in the fraction field which fails to be integral, then we will not have integrally closed, and hence we cannot possibly have UFD. $\endgroup$ – Daniel Montealegre Sep 9 '13 at 23:25
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    $\begingroup$ Read again the definition of "integrally closed". Your formulation is almost the converse of the correct one. Try to see if $1/2 \in \mathbb{Q}$ is integral over $\mathbb{Z}$ ... $\endgroup$ – Martin Brandenburg Sep 10 '13 at 7:51

It is an extremely subtle question to decide whether a domain is a UFD.
You can legally download this monograph by Samuel on the subject.

If you are geometrically minded the following criterion might illuminate the property of being a UFD:

A noetherian domain $D$ is a UFD if and only if it is integrally closed and its class group is trivial: $Cl(D)=0$

Notice that you may not replace the condition $Cl(D)=0$ by $Pic(D)=0$, as attested by the integrally closed domain $k[X,Y,Z]/(Z^2-XY)$ which has trivial Picard group but is not a UFD.


First of all, it should just be mentioned that sometimes there is some obvious element that does not have a unique factorization. The classic example of this is $6 \in \mathbb{Z}[\sqrt{-5}]$.

Secondly, there is the following fact. A Noetherian integral domain is a UFD if and only if all prime ideals of height $\leq 1$ are principal. Noetherian and integral domain are rather mild conditions, so this says that if you have a good grasp of the ideal structure, you can determine whether your ring is a UFD. An example of this being useful is in a Dedekind Domain. There you know that non-zero prime ideals are maximal and so all prime ideals have height less than or equal to 1. Moreover, by unique factorization of ideals, we can conclude that if all prime ideals are principal, then all ideals are principal. Thus, a Dedekind Domain being a UFD becomes equivalent to it being a PID.

  • $\begingroup$ If there is interest, I can provide a proof of the fact. $\endgroup$ – Alexander Sep 9 '13 at 23:51
  • $\begingroup$ What do you mean with height? $\endgroup$ – Daniel Montealegre Sep 10 '13 at 2:36
  • $\begingroup$ A prime ideal $P$ has height $n$ if the longest chain of prime ideals $P_1 \subset P_2 \subset ... \subset P$ has $n$ inclusions. Look at the wikipedia page on Krull dimension for more. So for instance, if all prime ideals are maximal, this says all primes are height $0$. If all nonzero primes are maximal, then $\{0\}$ is height $0$ and the rest are height $1$. $\endgroup$ – Alexander Sep 10 '13 at 2:45

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