Is there an easy example of an integral domain and two elements on it which do not have a greatest common divisor? It will have to be a non-UFD, obviously.

"Easy" means that I can explain it to my undergrad students, although I will be happy with any example.


5 Answers 5


Here's an example stolen blatantly from wikipedia.

Let $R=\mathbb{Z}[\sqrt{-3}]$, let $a=4=2*2=(1+\sqrt{-3})(1-\sqrt{-3})$ and $b=2(1+\sqrt{-3})$. Now, $2$ and $1+\sqrt{-3}$ are both maximal among divisors, but are not associates, thus, there is not GCD.

  • 1
    $\begingroup$ More generally, GCD domains are integrally closed. (Note that the real problem here is not about two elements, but one: the lattice of divisors of 4 does not have suprema.) $\endgroup$ Jan 8, 2010 at 5:26
  • 12
    $\begingroup$ Thanks. I should learn to look up wikipedia before asking. $\endgroup$ Jan 8, 2010 at 5:43

I should point out, there are plenty of examples in integrally closed rings. For example:

In $k[a,b,c,d]/(ad-bc)$, there is no GCD of $ad$ and $ab$. (Note that $a$ and $b$ are both common divisors.)

In $\mathbb{Z}[\sqrt{-5}]$, there is no GCD of $6$ and $2 (1+\sqrt{-5})$. (Note that $2$ and $1+\sqrt{-5}$ are both common divisors.)


It deserves to be much better known that nonexistant GCDs (and, similarly, nonprincipal ideals) arise immediately from any failure of Euclid's Lemma, and this provides an illuminating way to view many of the standard examples. Below is a detailed explanation extracted from one of my sci.math.research posts [2]. The results below hold true in any domain D.

LEMMA: (a,b) = (ac,bc)/c if (ac,bc) exists

Proof: d|a,b <=> dc|ac,bc <=> dc|(ac,bc) <=> d|(ac,bc)/c. QED

EUCLID'S LEMMA: a|bc and (a,b)=1 => a|c, if (ac,bc) exists

Proof: a|ac,bc => a|(ac,bc) = (a,b)c = c via Lemma. QED

Therefore if a,b,c fail to satisfy the implication in Euclid's Lemma, namely if (a,b) = 1 and a|bc, not a|c, then one immediately deduces that the gcd (ac,bc) fails to exist in D.

E.g. David Speyer's example above, and Khurana's example in [1] (= Theorem 41 in Pete L. Clark's [0]) are simply specializations where a,b,c = p,1+w,1-w in a quadratic number (sub)ring Z[w], ww = -d.

[0] Clark, Pete. L. Factorization in integral domains. 29pp. 2010. http://alpha.math.uga.edu/~pete/factorization2010.pdf

[1] D. Khurana, On GCD and LCM in domains: A Conjecture of Gauss. Resonance 8 (2003), 72-79. https://www.ias.ac.in/article/fulltext/reso/008/06/0072-0079

[2] sci.math.research, 3/12/09, seeking comments on expository article on factorization

  • $\begingroup$ Link [1] appears to be inoperative. $\endgroup$ Nov 6, 2020 at 6:35
  • $\begingroup$ @GerryMyerson: Updated the link. $\endgroup$ Jan 28, 2021 at 19:14
  • $\begingroup$ @darij, didn't work for me, but ias.ac.in/article/fulltext/reso/008/06/0072-0079 got me there. $\endgroup$ Jan 28, 2021 at 21:49
  • $\begingroup$ @GerryMyerson: OOps -- I misread "Link [1]" as "Link [0]" (which was inoperative, too) and fixed that one! $\endgroup$ Jan 28, 2021 at 21:56

Since Charles has already given you an example, I'll just mention that there is a name for integral domains in which any two non-zero elements have a gcd: GCD-Domains.

See also Pete's response to “Counter”-example for Gauss’s Lemma on irreducible polynomials.

  • $\begingroup$ I'd actually never heard of GCD-domains and GL-domains until this question, partly because I had missed that thread. Thanks Ben! $\endgroup$ Jan 8, 2010 at 5:37
  • $\begingroup$ Right, and in the Noetherian case a GCD-domain is precisely a UFD. So there are lots of examples. $\endgroup$ Jan 8, 2010 at 6:03

Another possible answer, following Qiaochu's comment: The elements $x^2$ and $x^3$ (Edit: $x^5$ and $x^6$) in $k[x^2, x^3]$ (alternatively, $k[x,y]/(x^3-y^2)$), where $k$ is a nonzero ring.

  • 1
    $\begingroup$ Non-integrally closed rings make my head spin, but don't those two elements have gcd 1? (I can't find any non-constant common divisors). I think x^4 and x^5 should work here, though. $\endgroup$ Jan 8, 2010 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.