Let $I$ be a nonzero ideal of the ring of Gaussian integers $\Z[i]$.

Prove that the quotient ring $\Z[i]/I$ is finite.

Proof.

Recall that the ring of Gaussian integers is a Euclidean Domain with respect to the norm
\[N(a+bi)=a^2+b^2\] for $a+bi\in \Z[i]$.
In particular, $\Z[i]$ is a Principal Ideal Domain (PID).

Since $I$ is a nonzero ideal of the PID $\Z[i]$, there exists a nonzero element $\alpha\in \Z[i]$ such that $I=(\alpha)$.
Let $a+bi+I$ be an arbitrary element in the quotient $\Z[i]/I$.
The Division Algorithm yields that
\[a+bi=q\alpha+r,\] for some $q, r\in \Z[i]$ and $N(r) < N(\alpha)$.

Since $a+bi-r=q\alpha \in I$, we have
\[a+bi+I=r+I.\] It follows that every element of $\Z[i]/I$ is represented by an element $r$ whose norm is less than $N(\alpha)$.

There are only finitely many elements in $\Z[i]$ whose norm is less than $N(\alpha)$.

(There are only finitely many integers $a, b$ satisfying $a^2+b^2 < N(\alpha)$.)

Hence the quotient ring $\Z[i]/I$ is finite.

The post The Quotient Ring $\Z[i]/I$ is Finite for a Nonzero Ideal of the Ring of Gaussian Integers first appeared on Problems in Mathematics.

Prove that the ring of integers
\[\Z[\sqrt{2}]=\{a+b\sqrt{2} \mid a, b \in \Z\}\] of the field $\Q(\sqrt{2})$ is a Euclidean Domain.

Proof.

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

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}).
\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.

For a proof of this problem, see the post “5 is prime but 7 is not prime in the ring $\Z[\sqrt{2}]$“.

The post The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain first appeared on Problems in Mathematics.