# How to show the quadratic integer ring O is not a UFD.

Let $$R=\mathbb{Z}[\sqrt{−n}]$$ where $$n$$ is a squarefree integer greater than 3. Prove that $$R$$ is not a UFD. Conclude that the quadratic integer ring O is not a UFD for $$D\equiv 2, 3$$ mod $$4$$, $$D < −3$$ (so also not an ED and not a PID). [Hint: Show that either $$\sqrt{−n}$$ or 1 + $$\sqrt{−n}$$ is not prime].

I have already shown that $$R$$ is not a UFD using the hint, but I am really stuck on how to conclude that the quadratic integer ring O is not a UFD for $$D\equiv 2, 3$$ mod $$4$$, $$D < −3$$.

Any guidance is appreciated! Thank you.

• What is $D$? If $D$ is just some negative integer and you're adjoining $\sqrt{D}$ to $\mathbb{Q}$ to obtain some number field, then the conclusion you're asked to draw is false even with the stated conditions on $D$. For example, take $D$ to be $-9$. Then $D \equiv 3 \pmod{4}$ but $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(i)$ and the ring of integers of this number field is a PID, hence is a UFD. Oct 20, 2022 at 0:39
• I guess you can try to use a representation $\sqrt{D}=d\sqrt{-n}$, where $d$ and $n$ are natural numbers and $n$ is squarefree. Sep 30, 2023 at 5:09

Lemma. Let $$R = \Bbb Z[\sqrt{-n}]$$ where $$n$$ is a squarefree integer greater than $$3$$. Then $$R$$ is not a UFD.
Theorem. Let $$K = \Bbb Q(\sqrt{M})$$, where $$M$$ is a squarefree integer (might be positive or negative) and $$\label{1}M \!\!\!\! \mod \!\!4 \in \{2, 3\}.$$ Then the integer ring $$O_K$$ is equal to $$\Bbb Z[\sqrt{M}]$$.
Thus, for a squarefree $$D$$ the statement holds, and for a non-squarefree $$D$$ it does not, as said in the comments.