# When UFD implies PID

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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If you remove UFD from the hypothesis, then it is satisfied by every Bezout domain. So, no, you can't remove it. – George Lowther Jun 10 '12 at 13:27
Any chance you could make that comment an answer? – Noah Snyder Jun 10 '12 at 20:43
@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? – Filippo Alberto Edoardo Jun 11 '12 at 3:49