MHB Is R a Principal Ideal Domain (P.I.D) Given These Properties?

[mathmari]

Hey!

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..
Since $R$ is Noetherian, we have that every ideal is finitely generated. Then every ideal is a finite product of irreducible elements of $R$.

How could we use the above property of $R$ to conclude that $R$ is a P.I.D. ? (Wondering)

 

[Deveno]

Suppose we have $I = (a,b)$.

Suppose $d = \gcd(a,b)$ (why does $d$ exist?).

Then $(a,b) = (ds,dt)$ where $\gcd(s,t) = 1$.

What can you say about $(d)$?

 

[mathmari]

Suppose we have $I = (a,b)$.

Suppose $d = \gcd(a,b)$ (why does $d$ exist?).

Then $(a,b) = (ds,dt)$ where $\gcd(s,t) = 1$.

What can you say about $(d)$?

Since $\gcd(s,t) = 1$ from the property of $R$ we have that $\exists u,v\in R$ such that $su+tv=1$.

Then $dsu+dtv=d \Rightarrow au+bv=d$.

We have that the elements of the ideal $(a,b)$ are of the form $ax+by$ where $x,y\in R$, right? (Wondering)

That means that elements of the ideal $(a,b)$ are the multiples of $d$, or not? ( $axk+byk=dk$ )
Therefore, $(a,b)=(d)$, right? (Wondering)

 

[Deveno]

Since $\gcd(s,t) = 1$ from the property of $R$ we have that $\exists u,v\in R$ such that $su+tv=1$.

Then $dsu+dtv=d \Rightarrow au+bv=d$.

This shows that $(d) \subseteq (a,b)$

We have that the elements of the ideal $(a,b)$ are of the form $ax+by$ where $x,y\in R$, right? (Wondering)

Yes.

That means that elements of the ideal $(a,b)$ are the multiples of $d$, or not? ( $axk+byk=dk$ )

Not quite right-your conclusion is true though.

We have $a = ds$ and $b = dt$, so $ax + by = (ds)x + (dt)y = d(sx + ty)$, so $(a,b) \subseteq (d)$

Therefore, $(a,b)=(d)$, right? (Wondering)

Now...can you use this to show $R$ is a PID?