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For simplicity, let $D$ be a square-free integer such that $D\equiv 2,3 \pmod 4$ such that the class number of $\mathbb{Z}[\sqrt{D}]$ is 1.

-Theorem in the page for "quadratic integer rings" in oeis

Let $p$ be a prime element of $\mathbb{Z}$.

If $p$ is not a prime element of $\mathbb{Z}[\sqrt{D}]$, then there exist two primes $\pi,\pi'$ in $\mathbb{Z}[\sqrt{D}]$ such that $p=\pi\pi'$.

How do I prove this? And is this theorem really a theorem? Actually I doubt this theorem.

Since $p$ is not prime in the quadratic integer ring, there are elements $a,b$ such that $p|ab$ and $p$ does not divide both $a$ and $b$. And I think this condition is too weak to imply the existence of such $\pi$ and $\pi'$.

I have rather proven that "if $p$ is not irreducible, then there are two irreducibles such that $p=\pi\pi'$". Moreover, I have shown that the above theorem is true in $\mathbb{Q}(\sqrt{D})$, but I doubt the case $\mathbb{Z}[\sqrt{D}]$.

If that theorem is really true, how do I prove it? Thank you in advance :)

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    $\begingroup$ $\pi$ and $\pi'$ are supposed to be ideals in this theorem. Then it is true. $\endgroup$
    – MooS
    Feb 28, 2015 at 21:13
  • $\begingroup$ @MooS I don't get you.. I wrote $\pi$ to mean an element, but what do you mean change an element to an ideal? $\endgroup$
    – Rubertos
    Feb 28, 2015 at 21:26
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    $\begingroup$ In the general case the theorem is indeed false, if $\pi, \pi'$ are supposed to be elements of the ring. I gave a counterexample in the comments below. $\endgroup$
    – MooS
    Feb 28, 2015 at 21:27
  • $\begingroup$ @MooS I get it. Thank you :) $\endgroup$
    – Rubertos
    Feb 28, 2015 at 21:29
  • $\begingroup$ Please edit the question to state that the ring is a UFD if this is what is intended (else it will confuse many future readers too). $\endgroup$
    – Bill Dubuque
    Feb 28, 2015 at 21:35

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I assume that $\mathbb Z[\sqrt D]$ is a UFD.

If $p$ isn't prime in $\mathbb Z[\sqrt D]$ there is $\pi\in\mathbb Z[\sqrt D]$ prime such that $p=\pi\pi'$, $\pi'\in\mathbb Z[\sqrt D]$. Now use the norm and find that $N(\pi')=\pm p$, so $\pi'$ is prime.

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    $\begingroup$ This is false. The theorem is not true, if $\pi$ is supposed to be an element of the ring. It has to be an ideal. Consider $3 \in \mathbb Z[\sqrt{-5}]$. $3$ is not prime, but there is no decomposition of the element $3$ into other elements. You have to decompose the ideal. $\endgroup$
    – MooS
    Feb 28, 2015 at 21:18
  • $\begingroup$ There is not a single reason to assume this. $\endgroup$
    – MooS
    Feb 28, 2015 at 21:19
  • $\begingroup$ @user26857 So you are assuming UFD to make $p$ reducible, which is what I proved already.. So the theorem stated itself is wrong? Moreover, I checked Theorem 9.29, but it is a theorem in $\mathbb{Q}(\sqrt{D})$, not the quadratic integer ring. $\endgroup$
    – Rubertos
    Feb 28, 2015 at 21:22
  • $\begingroup$ If the OP does only consider quadratic integer rings with class number $1$, he should really state this in his first question... $\endgroup$
    – MooS
    Feb 28, 2015 at 21:25
  • $\begingroup$ @MooS You are right, but I just realized that I misread the oeis article.. It was there assumed the rings to have class number 1. I'm sorry.. So I guess the statement in my question is in general false. $\endgroup$
    – Rubertos
    Feb 28, 2015 at 21:28

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