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

Question

My proof
$I=<a_1,\dots,a_n>=Da_1+Da_2+\cdots +Da_n$
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 $I=<g>$ 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

score 0 · Answer 1 · answered Nov 23 '19 at 16:35

0

Yes, the proof is correct.

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

answered Nov 23 '19 at 16:35

John Gowers

24,959

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

1 Answers1