# In which extensions of type $\mathbb{Q}(\sqrt{d})$ is the number 3 reducible?

Let $O$ be the ring of integers in $\mathbb Q(\sqrt{d})$ where $d$ is a square free negative number. Assume that the number 3 is reducible in $O$. Prove that $d = -2,-3,$ or $-11$.

Thank you for the help.

First there is problem to talk about norm where $d \lt -11$. There is a theorem that says when $d \lt -11$ the ring $\mathbb{Q}(\sqrt{d})$ isn't Euclidean. What can I do when $d \lt -11$ ?

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What do you mean reducible? Where the ideal $3$ splits? – Alex Youcis Feb 11 '14 at 10:15
@AlexYoucis reducible = not irreducible, I'd think. So there are (algebraic) integers $a,b$, neither a unit, with $a\cdot b = 3$. (Yes, of course that means the ideal splits, but the more elementary viewpoint of considering elements is probably more appropriate here.) – Daniel Fischer Feb 11 '14 at 10:28
Are you aware of the norm on $O$? – Daniel Fischer Feb 11 '14 at 10:32