A Prime Ideal in the Ring $\Z[\sqrt{10}]$

Prime Ideal Problems and Solution in Ring Theory in Mathematics

Problem 223

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 $\Z[\sqrt{10}]$.

 
LoadingAdd to solve later

Sponsored Links


Definition of a prime ideal.

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 $\Z[\sqrt{10}]$.

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


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

1 Response

  1. 06/27/2017

    […] 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 $Z[sqrt{10}]$“. […]

Please Login to Comment.

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

More in Ring theory
Prime Ideal Problems and Solution in Ring Theory in Mathematics
If a Prime Ideal Contains No Nonzero Zero Divisors, then the Ring is an Integral Domain

Let $R$ be a commutative ring. Suppose that $P$ is a prime ideal of $R$ containing no nonzero zero divisor....

Close