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I have a question as follows:

Does every pair of non-zero elements of $Z[\sqrt{3}]$ has a gcd?

Now, since $Z[\sqrt{3}]$ is a Euclidean domain as I saw a proof in one of the already answered questions on this site only, so it is a UFD. And in any UFD, any two non-zero elements have a gcd.
So, every pair of non-zero elements of $Z[\sqrt{3}]$ have a gcd.

Is my explanation correct? Can anyone kindly verify my answer?

Bernard
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Esha
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  • Right, for a proof see here. But perhaps you meant $R=\mathbb{Z}[\sqrt{-3}]$, where this is not the case. – Dietrich Burde Oct 19 '20 at 14:46
  • @Dietrich Burde No, I meant $Z[√3]$ only. Thanks for your verification. – Esha Oct 19 '20 at 14:54
  • Yes, a Euclidean domain is a gcd domain (i.e. all gcds exists). The proof is the same as in the classical proof in $\Bbb Z!:,$ any nonzero element $, d\in I = aR + bR,$ of least Euclidean value is a gcd of $,a,b,,$ since it divides all elements $,i\in I,$ (else $,0\neq i\bmod d\in I$ and its smaller than $d$, contra $d$ minimal). See the Remark here for more. – Bill Dubuque Oct 19 '20 at 21:44
  • Further a domain is UFD $\iff$ is a a gcd domain where divisibility is well-founded, i.e. there are no infinite divisibility chains $,\ldots d_3\mid d_2\mid d_1$ (equivalently it satisfies ACCP = ascending chain condition on principal ideals). – Bill Dubuque Oct 19 '20 at 21:54

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