Let $R$ be the subring $\big\{a+b\sqrt{10}~: a~,b\in{\bf Z}~\big\}$ of the field of real numbers.
The map $N:R\longrightarrow {\bf Z}$ given by $a+b\sqrt{10}\longmapsto(a+b\sqrt{10})(a-b\sqrt{10})=a^{2}-10b^{2}.$
Show that $u=0$ if $N(u)=0$ for $u\in R~,$where ${\bf Z}$ is the set of integers .
Here's my attempt :
Note that the ideal $(0)$ in this integral domain $R=\big\{a+b\sqrt{10}~: a~,b\in{\bf Z}~\big\}$ is prime ideal.
Then by assumption , we have $$(a+b\sqrt{10})(a-b\sqrt{10})=a^{2}-10b^{2}=N(u)=0\in (0)$$
But $(0)$ is prime ideal in this integral domain $R$ .
Therefore , $(a+b\sqrt{10})=0$ or $(a-b\sqrt{10})=0$ $\Longrightarrow a=-b\sqrt{10}$ or $a=b\sqrt{10}~~$ for $a,b\in{\bf Z}$
So , in either case $a=0=b,$ thus , $u=0~.$
Is there anyone that can check my work for validity ? Any suggestion or commend will be appreciated. Thanks for considering my request .