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The problem I was given: "If $m$ and $n$ are relatively prime rational integers, must they be relatively prime in every quadratic field $\mathbb{Q}[\sqrt d]$? If so, give a justification, if not, give a counterexample."

Where I'm stuck: How can integers be relatively prime over a quadratic field, this idea doesn't make sense to me as quadratic integers are of the form $a+b\times \sqrt d$

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  • $\begingroup$ Most probably they mean "in the ring of algebraic integers in the quadratic extension". This form of talk is widely used. $\endgroup$
    – DonAntonio
    Commented May 14, 2018 at 8:45

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Let $m,n \in \mathbb Z$. If $m$ and $n$ are relatively prime in $\mathbb Z$, then $am+bn=1$ for some $a,b \in \mathbb Z$.

Let $R$ be any subring of $\mathbb C$. Then $\mathbb Z \subseteq R$ and so $am+bn=1$ also holds in $R$.

Any number $\delta \in R$ that divides both $m$ and $n$ in $R$ must divide $1$ in $R$ and so is a unit.

This means that $m$ and $n$ are relatively prime in $R$.

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  • $\begingroup$ Nothing more needs to be said :-) $\endgroup$ Commented May 14, 2018 at 10:55

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