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Its not the case that every UFD is noetherian; the standard counterexample is $R[x_0,x_1,x_2,\ldots]$, which has the following ascending sequence of ideals:

$$\langle \rangle,\langle x_0\rangle,\langle x_0,x_1\rangle,\langle x_0,x_1,x_2\rangle,\ldots$$

But notice that most of these ideals aren't principal. So:

Question. Is it the case that every UFD is noetherian on principal ideals?

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    $\begingroup$ Yes, in fact "An integral domain $R$ is a UFD if and only if it satisfies ascending chain condition on principal ideals (ACCP) and every irreducible is prime." $\endgroup$ – Hamed Jun 16 '16 at 5:51
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    $\begingroup$ Yes. An integral domain is a UFD if and only if (1) principal ideals satisfy the ascending chain condition and (2) irreducible elements are prime. Basically (1) ensures the existence, (2) yields the uniqueness of factorizations. $\endgroup$ – Cihan Jun 16 '16 at 5:55
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I believe the answer is yes. Let $R$ be a UFD. Note that containment of two principal ideals $(a) \subset (b)$ corresponds with the division property $b \mid a$ for $a, b \in R$. Since $R$ is a UFD, every element has a unique finite factorization into irreducible elements, so let $a = x_{1}^{l_{1}}x_{2}^{l_{2}}\cdots x_{n}^{l_{n}}$ for irreducibles $x_{1}, \ldots, x_{n} \in R$ and some integers $l_{1}, \ldots, l_{n} \in \mathbb{N}$. Then $(a) \subset (b)$ implies that $b$ is a product of some proper subset of $\{x_{1}, \ldots, x_{n}\}$ raised to appropriate powers. It is then clear that every chain of principal ideals must eventually stabilize.

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