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 :)