Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Give an example of a UFD having a subring which is not a UFD.

I thought of $\mathbb{Z}[\sqrt{2},\sqrt{3}]$. Could you please explain my question. I am trying grasp the concepts, need help.

share|cite|improve this question
$\mathbb{Z}[\sqrt{-3}] \subset \mathbb{Z}[e^{2\pi i/3}]$. – Daniel Fischer Dec 20 '13 at 12:55
@DanielFischer I don't know if $\mathbb Z[e^{2\pi i / 3}]$ is a UFD ? – Complex analysis Dec 20 '13 at 13:04
@Complexanalysis It's a Euclidean ring. – Daniel Fischer Dec 20 '13 at 13:12
up vote 13 down vote accepted

Take any integral domain which is not a UFD and consider it as a subring of its field of fractions. (Fields are UFD for trivial reasons and if you don't accept this, take the polynomial ring over it)

share|cite|improve this answer
how can you be so fast... :D ...Simple. elegant.. – Praphulla Koushik Dec 20 '13 at 12:57

Consider $\mathbb C$ , it is a field hence obviously $UFD $ but if u consider a subring $\mathbb Z[\sqrt {-5}]$ is not a UFD . $9=3.3$ and also $9= (2+\sqrt{5} ) . (2-\sqrt5)$ , Hence the factorization is not unique.

share|cite|improve this answer

One question to ask after reviewing the two answers already posted is whether the existence of an extension $S\subset R$, with $R$ a UFD and the nonunits of $S$ being nonunits in $R$, makes $S$ a UFD.

While plausible, this is also false. For instance $k[x^2,x^3]\subset k[x]$ is a counterexample for any UFD $k$. This is because $x^2$ and $x^3$ are irreducible and yet $x^6=(x^2)^3=(x^3)^2$.

share|cite|improve this answer

Maybe I am beating the dead horse, but...

$\mathbb{Q} + X \mathbb{R}[X] \subset \mathbb{R}[X]$

is one of many similar examples.

share|cite|improve this answer

$\mathbb{Z}[2i]$ as a subset of $\mathbb{Z}[i]$. First one is not integrally closed, so can't be UFD, second is Euclidean.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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