Showing two elements are coprime in a ring of integers Let $\alpha$ and $\beta$ be the two roots of the polynomial $x^2 - x + 2$. I was wondering if someone could explain to me why $(y - \alpha)$ and $(y - \beta)$ are coprime (for any integer $y$) in the ring of integers of $\mathbb{Q}(\alpha)$?
Thanks!
 A: This is not true: $$4-\alpha=4-\frac{1+\sqrt{-7}}{2}=\frac{7-\sqrt{-7}}{2}=\sqrt{-7}(\frac{-1-\sqrt{-7}}{2}),$$ while $\frac{-1-\sqrt{-7}}{2}$ satisfies $x^2+x+2=0,$ and hence is an integer in $\mathbb Q(\alpha)$. Thus $\alpha-\beta=\sqrt{-7}$ divides both $4-\alpha$ and $4-\beta,$ while $\operatorname{N}(\alpha-\beta)=7$ shows that $\alpha-\beta$ is not a unit. So $4-\alpha$ and $4-\beta$ are not co-prime.  
In fact, we have the following  

Proposition
  $y-\alpha$ and $y-\beta$ are not co-prime if and only if $y\equiv4\pmod7,$ in which case $\gamma:=(\alpha-\beta)=\sqrt{-7}$ divides both of them.  

Proof
If $y\equiv4\pmod7,$ then $y-\alpha=4-\alpha+\sqrt{-7}(k\sqrt{-7})$ for some $k\in\mathbb Z,$ so $\gamma\mid y-\alpha;$ similarly, $\gamma\mid y-\beta.$
If $y\not\equiv4\pmod7,$ and if $\delta$ divides both $y-\alpha$ and $y-\beta,$ then $\delta\mid\gamma.$ As $\gamma$ is a prime in the ring $\mathbb Z[\alpha],$ either $\delta$ is a unit, or $\delta=\gamma.$
Since $\operatorname N(y-\alpha)=(y-\alpha)(y-\beta)=y^2-y+2\equiv(y-4)^2\pmod7,$ we find that $7=\operatorname N(\gamma)$ does not divide $\operatorname N(y-\alpha),$ thus $\gamma\not\mid(y-\alpha).$ Therefore $\delta$ is a unit, namely, every common divisor of $y-\alpha$ and $y-\beta$ is a unit, in other words, $y-\alpha$ and $y-\beta$ are coprime. $\square$
Hope this helps.
