# Let $D$ be a UFD such that Bezout's Identity holds. Then Every ideal is finitely generated implies that $D$ is PID [duplicate]

My proof
$$I==Da_1+Da_2+\cdots +Da_n$$
let $$g=gcd(a_1,\dots,a_n)$$
Then since Bezout's identity holds and the binary operator gcd is associative, $$g\in I$$.
Also any element of $$I$$ is divisible by $$g$$. Thus $$I=$$ and hence $$D$$ is PID

I have two questions, is the proof correct? and Where did I use the fact that $$D$$ is UFD? I think $$D$$ being Integral Domain with identity is enough as well.

• For the existance of a $\gcd$ in general you need a UFD, or more generally a GCD domain. – take008 Nov 23 '19 at 16:34
• @take008 A Bezout domain need not be a UFD, but it is always a GCD domain. – John Gowers Nov 23 '19 at 16:36
• @JohnGowers Ah, I see. I'm reading more about Bezout Domains now, thanks for the correction. – take008 Nov 23 '19 at 16:41

You don't need to use the fact that $$D$$ is a UFD: any Noetherian Bezout domain is a PID.