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

Problems and solutions of ring theory in abstract algebra

Problem 487

\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\] 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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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
\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}).

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

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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1 Response

  1. 06/27/2017

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

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