TITLE: $\mathbb{Z} [\sqrt{2}]$ is an integral domain
QUESTION [5 upvotes]: We know that $(\mathbb{Z} [\sqrt{2}],+,\cdot)$ is an integral domain.
Someone can prove it easily if he says that is a subring of $(\mathbb{R} ,+,\cdot)$ .
Can we find a different proof, more analytical? 
How can we show that $$\forall x,y\in \mathbb{Z} [\sqrt{2}], x\ne 0, y\ne 0\implies xy\neq 0$$

REPLY [8 votes]: In these cases, considering the conjugate can help. The conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$. Now, if $(a+b\sqrt{2})(c+d\sqrt{2})=0$, also
$$
(a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=0
$$
and therefore
$$
(a^2-2b^2)(c^2-2d^2)=0
$$
Since the integers form a domain, we conclude $a^2-2b^2=0$ or $c^2-2d^2=0$. The irrationality of $\sqrt{2}$ tells us that either $a=b=0$ or $c=d=0$.