## The Euclidean norm on ZZ[sqrt(-5)]

Let $N : \mathbb{Z}[\sqrt{\text{-}5}] \rightarrow \mathbb{N}$ be the Euclidean norm given by $N(a+b\sqrt{\text{-}5}) = a^2 + 5b^2$.

1. Show that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta$.
2. Show that $N(\alpha) = 1$ if and only if $\alpha$ is a unit.
3. Show that $N(\alpha) = 0$ if and only if $\alpha = 0$.
4. Show that if $N(\alpha)$ is prime as a natural number then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{\text{-}5}]$.
5. Show that $3 + 2\sqrt{\text{-}5}$ and $2 + \sqrt{\text{-}5}$ are prime.

Let $\alpha = a_1 + a_2\sqrt{\text{-}5}$ and $\beta = b_1 + b_2\sqrt{\text{-}5}$, and note the following.

 $N(\alpha\beta)$ = $N((a_1+a_2\sqrt{\text{-}5})(b_1+b_2\sqrt{\text{-}5}))$ = $N((a_1b_1 - 5a_2b_2) + (a_1b_2+a_2b_1)\sqrt{\text{-}5})$ = $(a_1b_1 - 5a_2b_2)^2 + 5(a_1b_2+a_2b_1)^2$ = $a_1^2b_1^2 + 5a_1^2 b_2^2 + 5a_2^2b_2^2 + 25 a_2^2b_2^2$ = $(a_1^2 + 5a_2^2)(b_1^2 + 5b_2^2)$ = $N(\alpha)N(\beta)$

Suppose $\alpha$ is a unit; then there exists $\beta$ such that $\alpha\beta = 1$. Considering norms, we have $N(\alpha)N(\beta) = 1$ (using the previous result). So $N(\alpha) = 1$. Conversely, suppose $N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 1$; then $b = 0$, and we have $a = \pm 1$. Thus $\alpha$ is a unit.

Certainly $N(0) = 0$. Now suppose $N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 0$; then $a = b = 0$.

If $N(\alpha)$ is prime and $\alpha = \beta\gamma$, then $N(\beta)N(\gamma)$ is prime. Without loss of generality, $N(\beta) = 1$, so that $\beta$ is a unit. Thus $\alpha$ is irreducible.

Note that $N(3+2\sqrt{\text{-}5}) = 3^2 + 5\cdot 2^2 = 29$ is prime; thus $3+2\sqrt{\text{-}5}$ is irreducible.

Note that $a^2 + 5b^2 = 3$ has no solutions in the integers since 3 is not a perfect square. In particular, no element of $\mathbb{Z}[\sqrt{\text{-}5}]$ has norm 3. Now $N(2 + \sqrt{\text{-}5}) = 9$. If $2 + \sqrt{\text{-}5} = \alpha\beta$, then $N(\alpha)N(\beta) = 9$. since no element of this ring has norm 3, without loss of generality we have $N(\alpha) = 1$ so that $\alpha$ is a unit. Thus $2 + \sqrt{\text{-}5}$ is irreducible.