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.

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
\[\pm 1, \pm i.\]

The Quotient Ring $\Z[i]/I$ is Finite for a Nonzero Ideal of the Ring of Gaussian Integers
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, […]

The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain
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 […]

Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]

A ring is Local if and only if the set of Non-Units is an Ideal
A ring is called local if it has a unique maximal ideal.
(a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$.
(b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$.
Prove that if every […]

5 is Prime But 7 is Not Prime in the Ring $\Z[\sqrt{2}]$
In the ring
\[\Z[\sqrt{2}]=\{a+\sqrt{2}b \mid a, b \in \Z\},\]
show that $5$ is a prime element but $7$ is not a prime element.
Hint.
An element $p$ in a ring $R$ is prime if $p$ is non zero, non unit element and whenever $p$ divide $ab$ for $a, b \in R$, then $p$ […]

Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal
Let $R$ be a commutative ring.
Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.
Proof.
$(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal
Suppose that $R$ is a field and let $I$ be a non zero ideal:
\[ \{0\} […]

How Many Solutions for $x+x=1$ in a Ring?
Is there a (not necessarily commutative) ring $R$ with $1$ such that the equation
\[x+x=1 \]
has more than one solutions $x\in R$?
Solution.
We claim that there is at most one solution $x$ in the ring $R$.
Suppose that we have two solutions $r, s \in R$. That is, we […]

The Quadratic Integer Ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD)
Prove that the quadratic integer ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD).
Proof.
Every element of the ring $\Z[\sqrt{5}]$ can be written as $a+b\sqrt{5}$ for some integers $a, b$.
The (field) norm $N$ of an element $a+b\sqrt{5}$ is […]