# Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$ ## 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? Add 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*}
&=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 $\Z[\sqrt{10}]$“. Add to solve later

### 1 Response

1. 06/27/2017

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

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

###### More in Ring theory ##### 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