Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$

Problems and solutions of ring theory in abstract algebra

Problem 487

Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\] be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\] Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?

 
LoadingAdd to solve later

Solution.

We prove that the ring $\Z[\sqrt{10}]/P$ is isomorphic to the ring $\Zmod{2}$.

We define the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ by sending $a+b\sqrt{10}$ to $\bar{a}=a \pmod 2 \in \Zmod{2}$.
The map $\Psi$ is a ring homomorphism. To see this,
let $a+b\sqrt{10}, c+d\sqrt{10} \in \Z[\sqrt{10}]$ .

We have
\begin{align*}
\Psi\left( (a+b\sqrt{10})(c+d\sqrt{10}) \right) &=\Psi\left(ac+10bd+(ad+bc)\sqrt{10}\right)\\
&=ac+10bd \pmod{2}=ac \pmod{2}\\
&=\Psi(a+b\sqrt{10}) \Psi(c+d\sqrt{10}).
\end{align*}

We also have
\begin{align*}
\Psi\left( (a+b\sqrt{10})+(c+d\sqrt{10}) \right) &=\Psi\left( a+c+(b+d)\sqrt{10}) \right) \\
&=a+c \pmod{2}\\
&=\Psi(a+b\sqrt{10})+\Psi(c+d\sqrt{10}).
\end{align*}

Therefore the map $\Psi$ is a ring homomorphism.

Since $\Psi(0)=\bar{0}$ and $\Psi(1)=\bar{1}$, the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is surjective.

We have $\Psi(a+b\sqrt{10})=\bar{0}$ if and only if $a$ is even.
Thus, the kernel of the homomorphism $\Psi$ is
\[\ker(\Psi)=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}=P.\]

In summary the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is a surjective ring homomorphism with the kernel $P$. Hence by the first isomorphism theorem, we have
\[\Z[\sqrt{10}] /P \cong \Zmod{2}\] as we claimed.

Since $\Zmod{2}$ is a field, the ideal $P$ is a maximal ideal, and in particular $P$ is a prime ideal.

Related Question.

A direct proof that the ideal $P=(2, \sqrt{10})$ is prime in the ring $\Z[\sqrt{10}]$ is given in the post “A prime ideal in the ring $\Z[\sqrt{10}]$“.


LoadingAdd to solve later

More from my site

You may also like...

1 Response

  1. 06/27/2017

    […] For a proof, see the post “Determine the Quotient Ring $Z[sqrt{10}]/(2, sqrt{10})$“. […]

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Your email address will not be published.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
Every Integral Domain Artinian Ring is a Field

Let $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring. Prove that...

Close