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It is well-known that the Bézout's lemma holds in any Euclidean domain. (The proof is based on Euclidean division.)

In some domains, the following special form still appears to be true despite being non-Euclidean: If $g=\gcd(a,b)$, then $\exists x,y: g=ax+by$.

How do you establish the above mentioned statement without referring to Euclidean division?

In particular, how do you prove it for $\mathbb{Z}[(1+\sqrt{-19})/2]$?

user26857
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    $\mathbb{Z}[\sqrt{-19}]$ is not normal, so there cannot be a Bézout lemma. Its integral closure is however a PID. See eg https://math.stackexchange.com/questions/246999/show-that-mathbbz-theta-where-theta-1-sqrt19i-2-is-a-princi. – Aphelli May 28 '22 at 13:28
  • How are you imagining Bezout be defined in a ring where there is no GCD functions? – Thomas Andrews May 28 '22 at 13:46
  • I mean, $5$ and $1+\sqrt{-19}$ have no common factor, but there is no solution to $5X+(1-\sqrt{-19})Y=1.$ Bezout is essentially the statement that all finitely-generated ideals are principal. – Thomas Andrews May 28 '22 at 13:52
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    Are you really lookng for the lemma in $\mathbb Z[\sqrt{-19}]$? Saying out loud what other commenters imply: the associated field $\mathbb Q[\sqrt{-19}]$ has integers other than those in $\mathbb Z[\sqrt{-19}]$, and you have to include those additional integers before you can have a PID and get the Bezout lemma. You need $\mathbb Z[(1+\sqrt{-19})/2]$. – Oscar Lanzi May 28 '22 at 15:32

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It is well known that your ring is a Principal Ideal Domain (for short, PID): Proof that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a PID which is not a Euclidean domain.

But a PID satisfies the Bezout lemma since every ideal is principal. In particular, for two elements $a,b$ we have $(a,b)=(d)$. It is not hard to show that $d=\gcd(a,b)$ and since $d\in(a,b)$ we can write $d=ax+by$.

user26857
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