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

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    $\begingroup$ Think of norms! $\endgroup$ Commented May 14, 2018 at 19:45
  • $\begingroup$ 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? $\endgroup$ Commented May 14, 2018 at 19:46
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    $\begingroup$ You don't need to think about norms; use Bezout's lemma. $\endgroup$ Commented May 14, 2018 at 19:47
  • $\begingroup$ or think of the fact that the ideal $(m,n)=(1)$ $\endgroup$
    – peter a g
    Commented May 14, 2018 at 19:47
  • $\begingroup$ we havent learned about Bezout's lemma yet $\endgroup$ Commented May 14, 2018 at 19:47

1 Answer 1

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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.

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  • $\begingroup$ this is the full solution? $\endgroup$ Commented May 14, 2018 at 19:59
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    $\begingroup$ @EricMaynard If you need to ask that then you probably need more study time on the basics... $\endgroup$
    – DonAntonio
    Commented May 14, 2018 at 20:00
  • $\begingroup$ BTW, +1.${}{}{}{}{}{}{}$ $\endgroup$
    – DonAntonio
    Commented May 14, 2018 at 20:00
  • $\begingroup$ I understand the solution @DonAntonio, but if you looked above, I asked for a hint or a solution so I was just making sure that I didn't miss something. $\endgroup$ Commented May 14, 2018 at 20:07

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