The prime number $p$ is prime in $\mathbb{Q}(\sqrt{d})$ iff $d$ is not a quadratic residue modulo p

Question

Theorem. The prime number $p$ is prime in $\mathbb{Q}(\sqrt{d})$ iff $d$ is not a quadratic residue modulo p.

Since this theorem has two directions, we have to prove two things. One way to do that is to assume that $d$ is quadratic residue, and to get the contradiction. My question is about the second part, about assuming that $p$ is not prime. I found the proof where it is said the following:

Assume that $p$ is not prime. Then, $p$ is not irreducible, so, there are $u,v \in \mathbb{Q}(\sqrt{d})$ such that $p=uv$ and $u,v$ are not unit elements. I won't continue writing this proof if it's not necessary, I have only one question: how come we concluded: $p$ is not prime $\Rightarrow$ $p$ is not irreducible? I know about the theorem that says that any prime in $\mathbb{Q}(\sqrt{d})$ is irreducible, but not vice versa. Can vice versa work in $\mathbb{Q}(\sqrt{d})$? Thank you.

Answer 1

$\Bbb{Q}(\sqrt{d})$ is a field it has no non-trivial (prime) ideal.

For $d \in \Bbb{Z}, \not \in \Bbb{Z}^2$ then $(p)$ is a prime ideal of $R=\Bbb{Z}[\sqrt{d}] =\Bbb{Z}[x]/(x^2-d)$ iff $R/(p) = \Bbb{F}_p[x]/(x^2-d)$ is an integral domain iff $x^2-d \in \Bbb{F}_p[x]$ is irreducible iff $d$ is a quadratic non-residue $\bmod p$.

With $d = m^2 D$ where $D$ is squarefree the ring of integers of $\Bbb{Q}(\sqrt{d})$ is $\Bbb{Z}[\sqrt{D}] $ or $\Bbb{Z}[\frac{1+\sqrt{D}}{2}] $, in the latter case there is the additional case of $(2)$ which is a prime ideal because $x^2+x+1 \in \Bbb{F}_2[x]$ is irreducible.