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

Unique Factorization Domain Problems and Solutions

Problem 519

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

 
LoadingAdd to solve later

Sponsored Links


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 defined by
\[N(a+b\sqrt{5})=(a+b\sqrt{5})(a-b\sqrt{5})=a^2-5b^2.\]

Consider the case when $a=3, b=1$.
Then we have
\[(3+\sqrt{5})(3-\sqrt{5})=4=2\cdot 2. \tag{*}\]

We prove that elements $2, 3\pm \sqrt{5}$ are irreducible in $\Z[\sqrt{5}]$.
Note that the norms of these elements are $4$.
We claim that each element $\alpha \in \Z[\sqrt{5}]$ of norm $4$ is irreducible.


Suppose that $\alpha=\beta \gamma$ for some $\beta, \gamma \in \Z[\sqrt{5}]$.
Our objective is to show that either $\beta$ or $\gamma$ is a unit.

Since we have
\[4=N(\alpha)=N(\beta \gamma)=N(\beta) N(\gamma)\] and the norms are integers, the value of $N(\beta)$ is one of $\pm 1, \pm 2, \pm 4$.

If $N(\beta)=\pm 1$, then $\beta$ is a unit.
If $N(\beta)=\pm 4$, then $N(\gamma)=\pm 1$ and hence $\gamma$ is a unit.

Let us consider the case $N(\beta)=\pm 2$.
We show that this case does not happen.
Write $\beta=a+b\sqrt{5}$ for some integers $a, b$.
Then we have
\[\pm 2 =N(\beta)=a^2-5b^2.\] Considering the above equality modulo $5$ yields that
\[\pm 2 \equiv a^2 \pmod{5}.\] However note that any square of an integer modulo $5$ is one of $0, 1, 4$.
So this shows that there is no such $a$.

Therefore, we have proved that either $\beta$ or $\gamma$ is a unit, hence $\alpha$ is irreducible.
The claim is proved.


It follows from (*) that the element $4 \in \Z[\sqrt{5}]$ has two different decompositions into irreducible elements.
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).

This problem only differs from the current problem by the sign.
($-5$ is used instead of $5$.)

For a proof of this problem, check out the post ↴
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. 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 […]
  • 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 […]
  • 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 $p$ […]
  • 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 […]
  • 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, […]
  • 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...

1 Response

  1. 07/25/2017

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

Please Login to Comment.

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

More in Ring theory
Unique Factorization Domain Problems and Solutions
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).

Close