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?
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Contents
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 $\Z[\sqrt{10}]$“.
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