MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
6  
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
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.