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

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

• Think of norms! Commented May 14, 2018 at 19:45
• i tried using contradictions with norms but i couldnt figure it out, do you mind writing out a partial-whole solution or just a hint? Commented May 14, 2018 at 19:46
• You don't need to think about norms; use Bezout's lemma. Commented May 14, 2018 at 19:47
• or think of the fact that the ideal $(m,n)=(1)$ Commented May 14, 2018 at 19:47
• we havent learned about Bezout's lemma yet Commented May 14, 2018 at 19:47

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.
• BTW, +1.${}{}{}{}{}{}{}$ Commented May 14, 2018 at 20:00