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.

  • $\begingroup$ Note: Prime implies irreducible, but in rings of the form $\mathbb{Q}(\sqrt{d})$, irreducible does not necessarily imply prime. A standard example is $1+\sqrt{-5}$ in $\mathbb{Z}(\sqrt{-5})$, since $2\times 3 = (1+\sqrt{-5})(1+\sqrt{-5})$, and all of the factors are irreducible in the ring of integers $\mathbb{Z}(\sqrt{-5})$. $\endgroup$ Aug 21, 2019 at 16:53
  • $\begingroup$ I thought the same, then the proof is not right? $\endgroup$
    – user672596
    Aug 21, 2019 at 16:54
  • $\begingroup$ P.S. I assume it means “in the ring of integers of $\mathbb{Q}(\sqrt{d})$”.... $\endgroup$ Aug 21, 2019 at 16:54
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    $\begingroup$ @karim-ashli: That’s related to my comment: in a field, every nonzero element is a unit, so nothing is a prime at all. So presumably the problem is actually asking about primes in the ring of integers of $\mathbb{Q}(\sqrt{d})$, rather than in $\mathbb{Q}(\sqrt{d})$ itself. $\endgroup$ Aug 21, 2019 at 17:01
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    $\begingroup$ (@Karim-ashli: Note that you are missing “is not zero and is not a unit” in your definition of prime). $\endgroup$ Aug 21, 2019 at 17:01

1 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.


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