Consider the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}\]
and its ideal
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}.\]
Show that $p$ is a prime ideal of the ring $\Z[\sqrt{10}]$.

An ideal $P$ of a ring $R$ is a prime ideal if whenever we have $ab \in P$ for elements $a, b \in R$, then either $a\in P$ or $b \in P$.

Proof.

Suppose that $a+b\sqrt{10}, c+d\sqrt{10} \in \Z[\sqrt{10}]$ and the product
\[(a+b\sqrt{10}) (c+d\sqrt{10}) \in P. \]
Then expanding the product, we have
\[ ac+10bd+(ad+bc)\sqrt{10} \in P.\]

Since $ac+10bd$ must be even number, we have either $a$ or $c$ is even.
Hence either
\[a+b\sqrt{10}\in P \text{ or } c+d\sqrt{10} \in P,\]
and we conclude that $P$ is a prime ideal.

Further Question.

In fact, it can be proved that $P$ is a maximal ideal in the ring $\Z[\sqrt{10}]$.

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[…] 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 $Z[sqrt{10}]$“. […]