# 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 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 at 10:28
Are you aware of the norm on $O$? –  Daniel Fischer Feb 11 at 10:32

We know (or maybe not) that the ring of integers in $\mathbb{Q}(\sqrt{d})$, where $d$ is not a perfect square, is $\mathbb{Z}[\omega_d]$, where

$$\omega_d = \begin{cases} \dfrac{1+\sqrt{d}}{2} &, d \equiv 1 \pmod{4}\\ \quad \sqrt{d} &, d \not\equiv 1\pmod{4}. \end{cases}$$

On all these rings, the norm of $\mathbb{Q}(\sqrt{d})$, $$N(\alpha + \beta\sqrt{d}) = (\alpha + \beta\sqrt{d})(\overline{\alpha + \beta\sqrt{d}}) = (\alpha + \beta\sqrt{d})(\alpha - \beta\sqrt{d}) = \alpha^2 - d\beta^2,$$ is integer valued, and in the basis $\{1,\omega_d\}$ it is expressed as

$$N(a + b\omega_d) = \begin{cases} a^2 + ab + \dfrac{1-d}{4}b^2 &, d \equiv 1 \pmod{4}\\ a^2 - db^2 &, d\not\equiv 1\pmod{4}. \end{cases}$$

The norm is multiplicative, $N(z\cdot w) = N(z)\cdot N(w)$, and $z \in \mathbb{Z}[\omega_d]$ is a unit if and only if $N(z) = \pm 1$.

Thus if $3$ is reducible, that is, there are non-units $z,w$ with $z\cdot w = 3$, then we have

$$N(3) = 3^2 = N(z)\cdot N(w),$$

where $\lvert N(z)\rvert > 1$ and $\lvert N(w)\rvert > 1$. By the unique prime factorisation in $\mathbb{Z}$, that implies

$$\lvert N(z)\rvert = \lvert N(w)\rvert = 3.$$

So a necessary condition for $3$ to be reducible is that there are ring elements $z$ with $N(z) = \pm 3$. That condition is also sufficient, since then $\pm z\cdot \overline{z} = \pm N(z) = 3$.

For $d < 0$, the norm only attains non-negative values, and we have

$$N(a + b\omega_d) = \begin{cases} a^2 + ab + \dfrac{1+\lvert d\rvert}{4}b^2 &, \lvert d\rvert \equiv 3 \pmod{4}\\ a^2 + \lvert d\rvert b^2 &, \lvert d\rvert\not\equiv 3\pmod{4} \end{cases}$$

in that case. From that, it is not difficult to derive necessary and sufficient conditions for $3$ to be reducible in $\mathbb{Z}[\omega_d]$ for $d < 0$.

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