Ring of Gaussian Integers and Determine its Unit Elements
Problem 188
Denote by $i$ the square root of $-1$.
Let
\[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\]
be the ring of Gaussian integers.
We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to
\[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]
Here $\bar{\alpha}$ is the complex conjugate of $\alpha$.
Then show that an element $\alpha \in R$ is a unit if and only if the norm $N(\alpha)=\pm 1$.
Also, determine all the units of the ring $R=\Z[i]$ of Gaussian integers.
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Proof.
Suppose that an element $\alpha$ is a unit of $R$.
Then there exists $\beta \in R$ such that $\alpha \beta=1$.
Then the norm of $\alpha \beta$ is
\begin{align*}
N(\alpha \beta)&=(\alpha \beta)(\overline{\alpha \beta})\\
&=\alpha \bar{\alpha} \beta \bar{\beta}\\
&=N(\alpha)N(\beta).
\end{align*}
Since the norm $N(1)=1$, we obtain
\[1=N(\alpha)N(\beta)\]
in the ring $\Z$. Since $N(\alpha)$ and $N(\beta)$ are both integers, it follows that we have
\[N(\alpha)=\pm 1 \text{ and } N(\beta)=\pm 1.\]
On the other hand, suppose that $N(\alpha)=\pm 1$ for an element $\alpha\in R$.
Then let $\beta:=N(\alpha)^{-1}\bar{\alpha}$.
Since $N(\alpha)^{-1}=\pm 1$, the element $\beta \in R$. We have
\begin{align*}
\beta \alpha&=N(\alpha)^{-1}\bar{\alpha}\cdot \alpha\\
&=N(\alpha)^{-1}N(\alpha)=1.
\end{align*}
Thus the element $\alpha$ is a unit in $R$.
Using this result, let us determine all units of the ring $R$ of Gaussian integers.
An element $\alpha=a+ib \in R$ is a unit if and only if
\[N(\alpha)=a^2+b^2=1,\]
where $a, b \in \Z$. Thus only solutions are
\[(a,b)=(\pm 1, 0), (0, \pm 1).\]
Therefore the units of $R=\Z[i]$ are
\[\pm 1, \pm i.\]
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