Are relatively prime rational integers still relatively prime in quadratic fields?

Question

If m and n are relatively prime rational (standard) integers, must they be relatively prime in every quadratic field Q[$\sqrt{d}$]?

Answer 1

Assume that $a+b\sqrt d$ divides both $n$ and $m$. Say, $(a+b\sqrt d)(x+y\sqrt d)=m$ and $(a+b\sqrt d)(u+v\sqrt d)=n$. As the irrational parts $(ay+bx)\sqrt d$, $(av+bu)\sqrt d$ must be zero, then also $(a-b\sqrt d)(x-y\sqrt d)=m$ and $(a-b\sqrt d)(u-v\sqrt d)=n$. After multiplication of the two variants, $(a^2-db^2)(x^2-dy^2)=m^2$ and $(a^2-db^2)(u^2-dv^2)=n^2$. As $m^2,n^2$ are also co-prime, we conclude $a^2-db^2=\pm1$, i.e., $a+b\sqrt d$ is a unit.