Ring of Gaussian Integers and Determine its Unit Elements

Problems and solutions of ring theory in abstract algebra

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.

 
LoadingAdd to solve later

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


LoadingAdd to solve later

Sponsored Links

More from my site

  • The Quotient Ring $\Z[i]/I$ is Finite for a Nonzero Ideal of the Ring of Gaussian IntegersThe 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 DomainThe 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 $1-ab$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 IdealA 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}]$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 IdealRing 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?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)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 […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Ring theory
Problems and solutions of ring theory in abstract algebra
$\sqrt[m]{2}$ is an Irrational Number

Prove that $\sqrt[m]{2}$ is an irrational number for any integer $m \geq 2$.  

Close