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The Bezout Lemma in the integers states that

For any $a, b \in\mathbb Z$, let $g = \gcd(a, b)$, There exists $x, y$ such that $ax+by = g$.

This can be generalized to a commutative ring that is a principal ideal domain (PID).

For any $a, b$ in a PID $R$, let $g = \gcd(a, b)$, which is defined by multiplying the common factors in their unique factorization (hence it is defined up to associates), then:

  1. There exists $x, y$ such that $ax+by = g$.
  2. The ideals generated by $a, b$ and $g$ respectively are the same: $(a, b) = (g)$

For the above setup, can we say $\gcd(x, y) = 1$? I think this is true (at least in the integers), but I got stuck when I tried to prove it.

My Proof Attempt: Say by contradiction that $\gcd(x, y) \ne 1$, then I can write $x = q\bar x, y = q\bar y$ for some non-unit $q$, then $$ax+by = g \Leftrightarrow q(a\bar x + b\bar y)=g$$ Obviously the element $a\bar x+b \bar y \in(a, b)=(g)$...

To complete the proof I need the fact that since $(a, b) = (g)$, $q$ has to be a unit. How can I do this? Thank you.

Ryan Lam
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  • From what you've written, it seems you want to prove that the gcd in a PID is always 1, which is not the case. Am I misreading something? – Andreas Caranti May 23 '22 at 11:32
  • See this proof in the linked dupe. – Bill Dubuque May 23 '22 at 11:44
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    I want to ask whether $gcd(x, y) = 1$, I think the above proof shows gcd(a, b) = 1? – Ryan Lam May 23 '22 at 12:29
  • @Ryan I don't think your question is a duplicate as none of the two questions above seem to answer it. I have an answer that I can't post now. But here's what you can do:

    because $a \bar x + b \bar y \in (g)$, it is equal to $cg$ for some $c$. Then we have $qcg=g$ and because we are in an integral domain, $qc=1$

     – Soheil Haghighi Jan 07 '25 at 19:38

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