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Is it true that an integral domain $R$ is a UFD if and only if intersection of any two principal ideals of $R$ is principal ?

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    $\begingroup$ To give a positive result: If you additionally require that every non-unit is a (finite) product of irreducible elements, then $R$ is a UFD. So your assumptions somehow give you the uniqueness of factorization once you have some factorization at first. $\endgroup$ – MooS Apr 14 '15 at 10:49
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A ring $R$ such that the intersection of any two principal ideals of $R$ is principal is exactly a GCD domain. And a GCD domain need not be an UFD.

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  • $\begingroup$ Could you please add a proof to the fact that an integral domain $R$ is a GCD domain if and only if intersection of any two principal ideals of $R$ is again principal ? $\endgroup$ – user228168 Apr 14 '15 at 13:01
  • $\begingroup$ @SaunDev This can be derived from the following two results: First, in an integer domain, $(a)\cap(b)$ is principal iff $\text{lcm}(a,b)$ exists, and in this case, $(a)\cap(b)=(\text{lcm}(a,b))$. Second, in an integer domain, if $\text{lcm}(a,b)$ exists, then $\gcd(a,b)$ exists and $\text{lcm}(a,b)\cdot\gcd(a,b)=ab$. Which results do you need a proof? $\endgroup$ – Censi LI Apr 14 '15 at 14:20
  • $\begingroup$ I need the proof that if $lcm$ exists then gcd exists . And what about the converse , Is it true that in a GCD domain , intersection of two principal ideals is principal ? $\endgroup$ – user228168 Apr 14 '15 at 14:29
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    $\begingroup$ @SaunDev If $\text{lcm}(a,b)$ exists, write it as $c$, it's sufficient to prove that $\frac{ab}c$ is the $\gcd$ of $a$ and $b$. Firstly, $\frac{ab}c$ divides $a$ and $b$ since $\frac{ab}c\cdot\frac cb=a$ and $\frac{ab}c\cdot\frac ca=b$. Secondly, if $d\mid a$ and $d\mid b$, then it's easy to see $\text{lcm}(\frac ad,\frac bd)=\frac cd$, which implies that $\frac cd\mid\frac ad\cdot\frac bd$, thus the conclusion. $\endgroup$ – Censi LI Apr 14 '15 at 16:13
  • $\begingroup$ @SaunDev Your conjecture that in a GCD domain, lcm always exists, is true. But the converse of the second of the above result fails, that is, in a integer domain, if for a certain pair of $a,b$, that $\gcd(a,b)$ exists doesn't imply that $\text{lcm}(a,b)$ exists. $\endgroup$ – Censi LI Apr 14 '15 at 16:16
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No, this is not true. A counterexample is the ring of holomorphic functions $R=Hol(\Bbb{C})$.

$R$ is not a UFD because irreducible (and prime as well) elements of $R$ are functions with exactly one zero with multiplicity 1. However, there are some functions with infinite zeroes (e.g. $\sin z$), and such functions cannot be factorized as a product of finitely many irreducible functions. A theorem by Weierstrass tells us that these functions can be written as an infinite product of irreducible functions. As an example $$\sin z = \pi z\prod_{k=-\infty, k \neq 0}^{+\infty} \left( 1 - \frac{z}{k} \right)$$

But now, suppose $f,g \in R$. Then $(f)\cap(g)$ is generated by a least common multiple of $f$ and $g$, which can be constructed factoring $f$ and $g$ as a (possibly infinite) product of irreducible functions.

This ring $R$ is not a UFD, but it is almost a UFD. Every element can be written as a (possibly infinite) product of irreducible elements. This can be done because of the topological structure of $\Bbb{C}$. In general, in an algebraic setting you are not allowed to write infinite products.

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