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Let $k=\mathbf Q[i]$ be the field of Gaussian numbers. I've proved the following easy lemma:

"If $x \in \mathcal{O}_k$ (the ring of integers of $k$) and $p$ is an odd rational prime dividing the norm $N_k(x)$, then there is a prime ideal $P$ in $\mathcal{O}_k$ such that $P$ divides both $p\mathcal{O}_k$ and $x\mathcal{O}_k$."

(The proof goes distinguishing if $p \equiv 1 \bmod 4$ or $p \equiv 3 \bmod 4$)

My question is if the same (or a similar) fact is true for other number fields $k$ (?)

Thanks for any idea/reference.

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    $\begingroup$ Doesn't it follow from the unique factorization into prime ideals? $\endgroup$
    – Fan Zheng
    Dec 12, 2016 at 18:24
  • $\begingroup$ My fault :-( I wanted to post the question on MSE not on MO. No problem if you want to move or close the question. $\endgroup$
    – sercej
    Dec 13, 2016 at 8:19

1 Answer 1

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Yes. It's sufficient to prove $(p,x)$ is not $(1)$, because then we can take any prime factor of $(p,x)$. But if $ap + bx =1 $ for $a,b \in \mathcal O_K$, then $N(b)N(x) =N(bx) = N(1-ap) \equiv 1$ mod $p$. This is a contradiction as $N(b)$ is an integer if $N(x)$ is divisible by $p$.

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  • $\begingroup$ A bit strange, to post an answer and vote to close as off-topic? $\endgroup$ Dec 12, 2016 at 22:08
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    $\begingroup$ @GerryMyerson A bit. Before I posted the answer there were no other off-topic votes, and one upvote. I thought maybe other people thought it was on-topic, and so answered it. $\endgroup$
    – Will Sawin
    Dec 12, 2016 at 22:34

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