A Bézout UFD is a PID.

Question

Let $R$ be an integral domain and a Noetherian U.F.D. with the following property:

for each couple $a,b\in R$ that are not both $0$, and that have no common prime divisor, there are elements $u,v\in R$ such that $au+bv=1$.

I want to show that $R$ is a P.I.D..

---

Could you give me some hints what we could do to show that $R$ is a P.I.D. ?

$$$$

EDIT:

Why is an ideal of the form $(x,y)$ principal?

Answer 1

Suppose $R$ is not a PID. Let $I$ be a non-zero non-principal ideal, and $0\ne a_1\in I$. Clearly $(a_1)\subsetneq I$. Let $b_2\in I\setminus (a_1)$. We have $(a_1)\subsetneq (a_1,b_2)\subsetneq I$ since the hypothesis tells us that the ideal $(a_1,b_2)$ is also principal. Set $(a_2)=(a_1,b_2)$, pick an element $b_3\in I\setminus(a_2)$, consider the ideal $(a_2,b_3)$ (which is also principal), and so on. This way you find a strictly ascending chain of principal ideals in $R$, a contradiction.

Answer 2

Every ideal generated by two elements is principal.

Let $a,b \in R$. Since $R$ is a UFD, we can write $a=da'$ and $b=db'$ with $d=\gcd(a,b)$ and $a',b'$ coprime. The hypothesis then imply that there are $u,v\in R$ such that $a'u+b'v=1$ and so $au+bv=d$. Therefore, $aR+bR=dR$, that is $(a,b)=(d)$.

Every finitely generated ideal is principal.

If $I=(r_1, r_2, \ldots, r_n)$ then by induction $J=(r_2, \ldots, r_n)$ is principal: $J=(s)$, and so $I=(r_1,s)$ is also principal by the previous fact.

Every ideal is principal.

Since $R$ is Noetherian, every ideal is finitely generated, and so is principal, by the previous fact.

All this can be summarized as:

The hypotheses imply that $R$ is a Bézout domain. Now a Bézout domain is a PID iff it is Noetherian.