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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.

Aleah Lillie
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    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. – user1090793 Oct 20 '22 at 00:39
  • 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. – Alex Ravsky Sep 30 '23 at 05:09

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Lemma. Let $R = \Bbb Z[\sqrt{-n}]$ where $n$ is a squarefree integer greater than $3$. Then $R$ is not a UFD.

Proof. See this answer and comments.

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 $\Bbb Z[\sqrt{M}]$.

Proof. See this, this or this (the proof structure is the same).

Thus, for a squarefree $D$ the statement holds, and for a non-squarefree $D$ it does not, as said in the comments.

Amateur_Algebraist
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