The Quotient Ring $\Z[i]/I$ is Finite for a Nonzero Ideal of the Ring of Gaussian Integers
Problem 534
Let $I$ be a nonzero ideal of the ring of Gaussian integers $\Z[i]$.
Prove that the quotient ring $\Z[i]/I$ is finite.
Add to solve laterSponsored Links
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.
Add to solve later
Sponsored Links
More from my site
![The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain](https://yutsumura.com/wp-content/uploads/2016/12/ring-theory-eye-catch-150x150.jpg)
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 […]- Ring of Gaussian Integers and Determine its Unit Elements
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 […]
Every Prime Ideal in a PID is Maximal / A Quotient of a PID by a Prime Ideal is a PID
(a) Prove that every prime ideal of a Principal Ideal Domain (PID) is a maximal ideal.
(b) Prove that a quotient ring of a PID by a prime ideal is a PID.
Proof.
(a) Prove that every PID is a maximal ideal.
Let $R$ be a Principal Ideal Domain (PID) and let $P$ […]- In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal
Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
Definition
A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element […]
- Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]
- Three Equivalent Conditions for an Ideal is Prime in a PID
Let $R$ be a principal ideal domain. Let $a\in R$ be a nonzero, non-unit element. Show that the following are equivalent.
(1) The ideal $(a)$ generated by $a$ is maximal.
(2) The ideal $(a)$ is prime.
(3) The element $a$ is irreducible.
Proof.
(1) $\implies$ […]
- Is the Quotient Ring of an Integral Domain still an Integral Domain?
Let $R$ be an integral domain and let $I$ be an ideal of $R$.
Is the quotient ring $R/I$ an integral domain?
Definition (Integral Domain).
Let $R$ be a commutative ring.
An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]
- Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals
Give an example of a commutative ring $R$ and a prime ideal $I$ of $R$ that is not a maximal ideal of $R$.
Solution.
We give several examples. The key facts are:
An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
An ideal $I$ of […]
