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Let $F$ be an algebraic Number Field and $O_F$ its ring of integers. If $\alpha \in O_F$ then is there a criterion for when $\alpha$, $\alpha'$ are relatively prime?\ \ I am most interested in $O_F=\mathbb{Z}[\zeta_3]$ for the time being.\ \ It seems like a possible criterion could be Let $a,b\in \mathbb{Z}$ and $\alpha=a+b\zeta$. Then, $gcd(a,b)=1$ iff $\alpha, \alpha'$ are relatively prime. Is this correct? The $\Rightarrow$ direction is most important to me.\ \ Thanks.\ \

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Maybe my pure ring theoretic number theory is a little rusty, but what is the general notion of GCD? In a Euclidean Domain you could take the set of common divisors and any that have largest norm would be GCD. But $\mathcal{O}_F$ is merely Dedekind, so does this notion make sense? You're fine for $\mathbb{Z}[\zeta_3]$, though. –  Matt Feb 28 '11 at 22:46
    
Yes, UFD is the only place this makes sense. –  Jason Smith Feb 28 '11 at 22:56
    
Also the concept of "its conjugate" is unclear in the general case: a number field of degree n has n different embeddings in $\Bbb C$ and an element has many conjugates. –  Andrea Mori Feb 28 '11 at 23:10
    
Yes that is true as well. Thanks. –  Jason Smith Feb 28 '11 at 23:33
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up vote 2 down vote accepted

In your case we have $\gcd(a + b \zeta, a + b \zeta^2) = \gcd(a + b \zeta, b(\zeta - 1))$. If $\gcd(a, b) = 1$ then this equals $\gcd(a + b \zeta, \zeta - 1) = \gcd(a + b, \zeta - 1)$ which equals $1$ if and only if $3 \nmid a + b$.

In general I am not sure what you mean by "relatively prime" if $\mathcal{O}_F$ is not a UFD. Do you mean that they generate the unit ideal? There are lots of ways to check this generalizing the method above. A sufficient condition is that $(\alpha, \alpha') = (x, y)$ where the norms of $x, y$ are relatively prime.

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