Looking for an example of a GCD domain which is not a UFD

Question

I know that every UFD (unique factorization domain) is a GCD domain i.e. g.c.d. of any two elements, not both zero, exists in the domain.

I am looking for an example of a GCD domain which is not a UFD.

I have not been able to find mainly for the difficulty that in a GCD domain every irreducible must be a prime and all the examples of non-UFD's I know, in them some irreducible is not prime.

So please help. Thanks in advance.

Fore some examples see here. – user26857 May 04 '15 at 10:46 — user26857, May 04 '15 at 10:46

score 14 · Answer 1 · edited Apr 13 '17 at 12:20

14

Let $\mathcal{O}$ be the ring of all algebraic integers. Then $\mathcal{O}$ is a GCD domain as $\mathcal{O}$ is a Bezout Domain [see] and all Bezout domains are GCD domains, but $\mathcal{O}$ is not UFD as it is not even a factorization domain as it has no irreducible elements as for any $a\in \mathcal{O}$, $a=\sqrt{a}\cdot\sqrt{a}$.

edited Apr 13 '17 at 12:20

Community

1

answered Aug 18 '15 at 23:34

Bhaskar Vashishth

11,279

score 7 · Answer 2 · edited Dec 15 '18 at 07:20

7

I know two interesting examples of GCD domains which are not UFDs (in fact, both are Bézout domains). The first is the ring of all entire functions over $\mathbb C$; the second is the monoid ring $K\mathbb Q^+$, where $K$ denotes an arbitrary field, and $\mathbb Q^+$ denotes the (additive) monoid of all non negative rational numbers.

For the definition of monoid ring, see wiki. Roughly speaking, $K\mathbb Q^+$ is a generalization of polynomial rings, you can view it as the ring of all "polynomials" with non negative rational exponents (instead of nonnegative integral exponents).

edited Dec 15 '18 at 07:20

user26857

52,094

answered Apr 11 '15 at 13:14

Censi LI

5,905

Could you please elaborate on monoid ring ? I mean what are the operations of $K \mathbb Q^+ $ ? – Apr 11 '15 at 13:22
What is the gcd of two entire functions? – Nishant Apr 11 '15 at 13:23
@SaunDev I've added a bit more details, do you understand it now? – Censi LI Apr 11 '15 at 13:29
3

@Nishant $\gcd(f,g)$ is any entire function of which the set of zeros is exactle the common zeros of $f$ and $g$, and the multiplicity of each zero is the minimum of that of $f$ and $g$. – Censi LI Apr 11 '15 at 13:31
1

@SaunDev For details on the second example see here. – Bill Dubuque Apr 14 '15 at 19:44

Pedro · Answer 3 · 2015-08-19T00:22:37.903

An example is the ring of holomorphic functions $\mathcal O(\Bbb C)$. This is a Bézout domain, so a GCD domain. It is not a UFD since the irreducible elements are linear polynomials, but there are holomorphic functions with infinitely many roots. Since $\mathcal O(\Bbb C)$ is not noetherian (for example, the ideal that of functions that vanish on a finite subset of $\Bbb Z$ is not finitely generated), so it is not a PID (this can be deduced from the fact it is not a UFD, however).

Add I realize this has given as an example above. In fact

Wedderburn's Lemma Given $f,g\in \mathcal O(\Bbb C)$ that are relatively prime there always exist $h,k\in \mathcal O(\Bbb C)$ such that $fh+gk=1$. Thus given finitely many entire $f_1,\ldots,f_r$ and a gcd $f$, there always exist entire $g_i$ such that $f_1g_1+\dots+f_rg_r=f$.

Looking for an example of a GCD domain which is not a UFD

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