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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.

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  • $\begingroup$ Fore some examples see here. $\endgroup$ – user26857 May 4 '15 at 10:46
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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 $a\in \mathcal{O}$, $a=\sqrt{a}\cdot\sqrt{a}$.

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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).

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  • $\begingroup$ Could you please elaborate on monoid ring ? I mean what are the operations of $K \mathbb Q^+ $ ? $\endgroup$ – user228168 Apr 11 '15 at 13:22
  • $\begingroup$ What is the gcd of two entire functions? $\endgroup$ – Nishant Apr 11 '15 at 13:23
  • $\begingroup$ @SaunDev I've added a bit more details, do you understand it now? $\endgroup$ – Censi LI Apr 11 '15 at 13:29
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    $\begingroup$ @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 $g$. $\endgroup$ – Censi LI Apr 11 '15 at 13:31
  • $\begingroup$ @SaunDev For details on the second example see here. $\endgroup$ – Bill Dubuque Apr 14 '15 at 19:44
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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 $f_1g_1+\dots+f_rg_r=f$.

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