# Prime element of a quadratic integer ring

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 :)

• $\pi$ and $\pi'$ are supposed to be ideals in this theorem. Then it is true.
– MooS
Feb 28, 2015 at 21:13
• @MooS I don't get you.. I wrote $\pi$ to mean an element, but what do you mean change an element to an ideal? Feb 28, 2015 at 21:26
• 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.
– MooS
Feb 28, 2015 at 21:27
• @MooS I get it. Thank you :) Feb 28, 2015 at 21:29
• 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). Feb 28, 2015 at 21:35

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.

• 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.
– MooS
Feb 28, 2015 at 21:18
• There is not a single reason to assume this.
– MooS
Feb 28, 2015 at 21:19
• @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. Feb 28, 2015 at 21:22
• If the OP does only consider quadratic integer rings with class number $1$, he should really state this in his first question...
– MooS
Feb 28, 2015 at 21:25
• @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. Feb 28, 2015 at 21:28