The following result is too elementary, both to state and to prove, not to be known. Can someone give a reference? Is there any hope if you don't suppose UFD (i.e. move that from the hypothesis to the conclusion)?

Theorem. Let $R$ be a commutative UFD with field of fractions $F$. Suppose that for any subring $S$ of $F$ that properly includes $R$, there is some non-unit of $R$ that becomes invertible in $S$. Then $R$ is a PID.

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    $\begingroup$ If you remove UFD from the hypothesis, then it is satisfied by every Bezout domain. So, no, you can't remove it. $\endgroup$ Jun 10, 2012 at 13:27
  • $\begingroup$ Any chance you could make that comment an answer? $\endgroup$ Jun 10, 2012 at 20:43
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    $\begingroup$ @George: The distinction PID vs Bézout is only in noetheriannes, right So MO's question may be changed by replacing "UFD" with "noetherian domain". Right? $\endgroup$ Jun 11, 2012 at 3:49


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