The Quadratic Integer Ring $\Z[\sqrt{-5}]$ is not a Unique Factorization Domain (UFD)

Unique Factorization Domain Problems and Solutions

Problem 518

Prove that the quadratic integer ring $\Z[\sqrt{-5}]$ is not a Unique Factorization Domain (UFD).

LoadingAdd to solve later

Sponsored Links


Proof.

Any element of the ring $\Z[\sqrt{-5}]$ is of the form $a+b\sqrt{-5}$ for some integers $a, b$.
The associated (field) norm $N$ is given by
\[N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2.\]

Consider the case when $a=2, b=1$.
Then we have
\begin{align*}
(2+\sqrt{-5})(2-\sqrt{-5})=9=3\cdot 3. \tag{*}
\end{align*}

We claim that the numbers $3, 2\pm \sqrt{-5}$ are irreducible elements in the ring $\Z[\sqrt{-5}]$.

To prove the claim at once, we show that any element in $\Z[\sqrt{-5}]$ of norm $9$ is irreducible.


Let $\alpha$ be an element in $\Z[\sqrt{-5}]$ such that $N(\alpha)=9$.
Suppose that $\alpha=\beta \gamma$ for some $\beta, \gamma \in \Z[\sqrt{-5}]$.
Out goal is to show that either $\beta$ or $\gamma$ is a unit.

We have
\begin{align*}
9&=N(\alpha)=N(\beta)N(\gamma).
\end{align*}
Since the norms are nonnegative integers, $N(\beta)$ is one of $1, 3, 9$.

If $N(\beta)=1$, then it yields that $\beta$ is a unit.

If $N(\beta)=3$, then we write $\beta=a+b\sqrt{-5}$ for some integers $a, b$, and we obtain
\[3=N(\beta)=a^2+5b^2.\] A quick inspection yields that there are no integers $a, b$ satisfying this equality.
Thus $N(\beta)=3$ is impossible.

If $N(\beta)=9$, then $N(\gamma)=1$ and thus $\gamma$ is a unit.

Therefore, we have shown that either $\beta$ or $\gamma$ is a unit.

Note that the elements $3, 2\pm \sqrt{-5}$ have norm $9$, and hence they are irreducible by what we have just proved.


It follows from the equalities in (*) that the factorization of the element $9$ into irreducible elements are not unique.
Thus, the ring $\Z[\sqrt{-5}]$ is not a UFD.

Related Question.

Problem.
Prove that the quadratic integer ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD).

Note that $-5$ is replaced by $5$.

See the proof of this problem ↴
The Quadratic Integer Ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD)


LoadingAdd to solve later

Sponsored Links

More from my site

  • 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 […]
  • The Ideal Generated by a Non-Unit Irreducible Element in a PID is MaximalThe Ideal Generated by a Non-Unit Irreducible Element in a PID is Maximal Let $R$ be a principal ideal domain (PID). Let $a\in R$ be a non-unit irreducible element. Then show that the ideal $(a)$ generated by the element $a$ is a maximal ideal.   Proof. Suppose that we have an ideal $I$ of $R$ such that \[(a) \subset I \subset […]
  • Ring of Gaussian Integers and Determine its Unit ElementsRing 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 […]
  • 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 […]
  • 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 […]
  • 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, […]
  • 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 $p$ […]
  • Three Equivalent Conditions for an Ideal is Prime in a PIDThree 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) $\implies$ […]

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
Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$

Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$. Let $(x)$ be the...

Close