First of all, it is clear that $\Z[\sqrt{2}]$ is an integral domain since it is contained in $\R$.

We use the norm given by the absolute value of field norm.
Namely, for each element $a+\sqrt{2}b\in \Z[\sqrt{2}]$, define
\[N(a+\sqrt{2}b)=|a^2-2b^2|.\]
Then the map $N:\Z[\sqrt{2}] \to \Z_{\geq 0}$ is a norm on $\Z[\sqrt{2}]$.
Also, it is multiplicative:
\[N(xy)=N(x)N(y).\]
Remark that since this norm comes from the field norm of $\Q(\sqrt{2})$, the multiplicativity of $N$ holds for $x, y \in \Q(\sqrt{2})$ as well.

We show the existence of a Division Algorithm as follows.
Let
\[x=a+b\sqrt{2} \text{ and } y=c+d\sqrt{2}\]
be arbitrary elements in $\Z[\sqrt{2}]$, where $a,b,c,d\in \Z$.

We have
\begin{align*}
\frac{x}{y}=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\frac{(ac-2bd)+(bc-ad)\sqrt{2}}{c^2-2d^2}=r+s\sqrt{2},
\end{align*}
where we put
\[r=\frac{ac-2bd}{c^2-2d^2} \text{ and } s=\frac{bc-ad}{c^2-2d^2}.\]

Let $m$ be an integer closest to the rational number $r$ and let $n$ be an integer closest to the rational number $s$, so that
\[|r-m| \leq \frac{1}{2} \text{ and } |s-n| \leq \frac{1}{2}.\]

Let
\[t:=r-n+(s-m)\sqrt{2}.\]

Then we have
\begin{align*}
t&=r+s\sqrt{2}-(n+m\sqrt{2})\\
&=\frac{x}{y}-(n+m\sqrt{2})y.
\end{align*}

It follows that
\begin{align*}
yt=x-(n+m\sqrt{2})y \in \Z[\sqrt{2}].
\end{align*}

Thus we have
\begin{align*}
x=(n+m\sqrt{2})y+yt \tag{*}
\end{align*}
with $n+m\sqrt{2}, yt\in \Z[\sqrt{2}]$.

We have
\begin{align*}
N(t)&= |(r-n)^2-2(s-m)^2|\\
&\leq |r-n|^2+2|s-m|^2\\
& \leq \frac{1}{4}+2\cdot\frac{1}{4}=\frac{3}{4}.
\end{align*}

It follows from the multiplicativity of the norm $N$ that
\begin{align*}
N(yt)=N(y)N(t)\leq \frac{3}{4}N(y)< N(y).
\end{align*}
Thus the expression (*) gives a Division Algorithm with quotient $n+m\sqrt{2}$ and remainder $yt$.

Related Question.

Problem. In the ring $\Z[\sqrt{2}]$, prove that $5$ is a prime element but $7$ is not a prime element.

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## 1 Response

[…] For a proof of this fact, see that post “The Ring $Z[sqrt{2}]$ is a Euclidean Domain“. […]