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 […]

The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic
Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.
Proof.
We give three proofs.
The first two proofs use only the properties of ring homomorphism.
The third proof resort to the units of rings.
If you are familiar with units of $\Z[x]$, then the […]

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 […]