Is it Integral domain?

Question

Is $ \mathbb{Z}[\sqrt {D}]$ an integral domain ?

My solution:

$$ \mathbb{Z}[\sqrt {D}]=\{ a+b\sqrt {D} , a,b \in \mathbb{Z}\}$$

No ,

Suppose $\mathbb {Z}[\sqrt {D}]$ is integral domain then Let $a, b \in \mathbb {Z}[\sqrt {D}]$ Where$ a=x +y\sqrt {D}$

$||a||$ $||b||=0$
$ \implies ||a|| =0 or ||b||=0$

If $||a|| =0 $then $x^2-y^2D=0$ So $x= y\sqrt {D}$

then it is contradiction when D is not perfect square , so it isn't Integra domain.

If D is prefect square then it is integral domain

Can anyone tell me if it is true or not? if not , any hint for solve ? Thanks

Answer 1

Hint:

Observe that no matter what $\;D\in\Bbb Q\;$ is, we always have $\;\Bbb Z(\sqrt D)\subset \Bbb C\;$ ...