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I'm studying $\mathbb{Z}[i]$, Gauss' integers, which I know is an euclidean ring. I want to prove that if $x,y\in\mathbb{Z}$ are such that $x\wedge y=1$, one being odd and the other even, and $\omega = x+iy\in\mathbb{Z}[i]$, then $\omega\wedge \overline{\omega}=1$ in $\mathbb{Z}[i]$.

I've started with $\omega\wedge \overline{\omega} = (x+iy)\wedge(2x)$, I don't know what to do now.

I notated $x\wedge y = \mathrm{gcd}(x,y)$.

Rafaël
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    how about the case $x=y=1$? – user8268 May 03 '23 at 13:58
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    If $a+ib$ divides $\omega$ and $\bar\omega$ then $a^2+b^2\mid\gcd(4x^2,4y^2)=4.$ It is then easy to conclude. – Anne Bauval May 03 '23 at 14:04
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    Building on the comment of @user8268, if $z$ divides both $\omega$ and $\overline{\omega}$, then it divides $2 x = \omega + \overline{\omega}$ and $2 y = i (\overline{\omega} - \omega)$, so that $z$ divides $2$. Up to units, this leaves the cases $z =1, 1+ i, 2$. Clearly $z = 2$ is impossible, as $x, y$ are not both even, but $1 + i$ is well possible, when $x$ and $y$ are both odd (and coprime). – Andreas Caranti May 03 '23 at 14:14
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    Pleased define the nonstandard notation $, a\wedge b\ \ $ – Bill Dubuque May 03 '23 at 17:50
  • I don't think I understand one step in your solutions: I get that $z$ divides $\omega$ and $\overline{\omega}$ implies $z$ divides $2x$ and $2y$ and thus $z$ divides gcd$(2x,2y)$, but why is this gcd (in $\mathbb{Z}[i]$) the same as the gcd in $\mathbb{Z}$ (which I agree is indeed $2$ here) ? – Rafaël May 04 '23 at 09:39
  • @Rafaël See here for why such gcds persist in extensions (more generally). – Bill Dubuque May 08 '23 at 17:53

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