TITLE: Is $\mathbb{Z}[\sqrt{15}]$ a UFD?
QUESTION [13 upvotes]: Let $R=\mathbb{Z}[\sqrt{15}]=\{a+b\sqrt{15}:a,b\in\mathbb{Z}\}$.


How do I show that $(3,\sqrt{15})$ is a maximal ideal but not a principal ideal?  
How do I show that $(3,\sqrt{15})^2$ is a principal ideal?  
How do I show that $R$ is (not) a UFD?


What I have done:  

If $(3,\sqrt{15})$ is a maximal ideal, then I must show that $R/(3,\sqrt{15})$ is a field. I thought that this holds:$R/(3,\sqrt{15})=\mathbb{Z}/3\mathbb{Z}$. Is this correct? How do I proceed from here?  
I know that $(3,\sqrt{15})^2=(9,3\sqrt{15},15)$. How can I use this?  
I'm afraid that I don't know where to start with this one. Maybe one of the statements above can help?

Thanks for taking the time!

REPLY [7 votes]: $$\mathbb Z[\sqrt{15}]\simeq\mathbb Z[X]/(X^2-15),$$
so
$$\mathbb Z[\sqrt{15}]/(3,\sqrt{15})\simeq\frac{\mathbb Z[X]/(X^2-15)}{(3,X,X^2-15)/(X^2-15)}\simeq\mathbb Z[X]/(3,X)\simeq\mathbb Z/3\mathbb Z.$$ Thus $(3,\sqrt{15})$ is a maximal ideal.