Prove that $\alpha\in \mathbb{Z}[\sqrt{2}]^*$ if and only if $N(\alpha)=\pm 1$.

Question

How would I show that an element $\alpha$ is in $\mathbb{Z}[\sqrt{2}]^*$ (the invertible elements of $\mathbb{Z}[\sqrt{2}]$) if and only if the norm of $\alpha$ is $\pm 1$? I did not see this as a duplicate at first because I thought that $1$ and $\pm 1$ needed different ways of proving, but it turns out that if $N(\alpha)$ divides $1$, then it can be negative or positive.

Answer 1

The norm of $\alpha=a+b\sqrt{2}$ is given by $N(\alpha)=\alpha \overline{\alpha}$, where $\overline{\alpha}=a-b\sqrt{2}$.

If $N(\alpha)=\pm1$, then $\alpha \overline{\alpha}=\pm1$, so $\alpha\mid 1$, which means that $\alpha$ is invertible.

Now, if $\alpha$ is invertible, then $\alpha^{-1}$ exists and $\alpha \alpha^{-1}=1$, which lead us to $$N(\alpha \alpha^{-1})=N(\alpha)N(\alpha^{-1})=N(1)=1.$$

Hence $N(\alpha)\mid 1$, which implies that $N(\alpha)=\pm1$.

Answer 2

First off all, consider the map

$$c:\mathbb{Z}[\sqrt{2}]\to \mathbb{Z}[\sqrt{2}],\,a+b\sqrt{2}\mapsto a-b\sqrt{2}$$

This map is well defined because of the uniqueness of representation of elements in $\mathbb{Z}[\sqrt{2}]$ in the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Z}$ (which is in turn a consequence of the irrationality of $\sqrt{2}$).

Then consider the map :

$$N:\mathbb{Z}[\sqrt{2}]\to \mathbb{Z}[\sqrt{2}],\,u\mapsto u\,c(u)$$

and show that $N$ is multiplicative, which means that :

$$\forall(u,v)\in \mathbb{Z}[\sqrt{2}]^2,\,N(uv)=N(u)N(v)$$

Now, let $a+b\sqrt{2}$ be an invertible element. Then there exists $c+d\sqrt{2}$ such that :

$$(a+b\sqrt{2})(c+d\sqrt{2})=1$$

Applying $N$ to both sides and using the fact that $N$ is multiplicative, we get :

$$N(a+b\sqrt{2})N(c+d\sqrt{2})=N(1)$$

that is :

$$(a^2-2b^2)(c^2-2d^2)=1$$

$a^2-2b^2$ and $c^2-2d^2$ are integers and their product is $1$, such each of them must be $\pm 1$