$\mathbb{Z} [\sqrt{2}]$ is an integral domain 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 $$\forall x,y\in \mathbb{Z} [\sqrt{2}], x\ne 0, y\ne 0\implies xy\neq 0$$
 A: 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 $c=d=0$.
A: The main proof you mention is the easiest and the best since it generalizes very well. If you really want something more intrinsic, note
$$(a+b\sqrt 2)(c+d\sqrt 2)=0\implies (a^2-2b^2)(c^2-2d^2)=0$$
But then if so, either $a^2=2b^2$ or $c^2=2d^2$, WLOG assume the former.
Then $a$ is even, but then if the prime factorization of $a$ is $2^kp_1^{e_1}\ldots p_r^{e_r}$ we have

$$a^2=2^{2k}p_1^{2e_1}\ldots p_r^{2e_r}$$

So the exponent of $2$ in $a$ is even, however since $b^2$ has an even power of $2$, $2b^2$ has an odd power, a contradiction, unless that power is infinity, but the only number infinitely divisible by $2$ is $0$, so $a=b=0$ holds
A: Hint Suppose we have $$(a + b \sqrt{2})(c + d \sqrt{2}) = 0$$ from some $a, b, c, d \in \Bbb Z$. Expanding gives $$(ac + 2 bd) + (ad + bc) \sqrt{2} = 0,$$
and since $\sqrt{2}$ is irrational, the coefficients must vanish separately
$$ac + 2bd = ad + bc = 0 .$$
Substituting $0$ for any of the four parameters gives quickly that $a = b = c = d = 0$, so we may assume that none is zero. Then, it follows from the second equation that $c = -\lambda a$ and $d = \lambda b$ for some $\lambda \in \Bbb Q - \{ 0 \}$.

Substituting gives $$0 = a c + 2 b d = a(-\lambda a) + 2 b (\lambda b) = \lambda (-a^2 + 2b^2) .$$ Clearing $\lambda$ and rearranging gives $\left(\frac{a}{b}\right)^2 = 2$, but this contradicts the irrationality of $\sqrt{2}$.

