Show that $p(x)=x^2-\sqrt{2}$ is irreductible on $\mathbb{Z}[\sqrt{2}]$

Question

Show that $p(x)=x^2-\sqrt{2}$ is irreductible on $\mathbb{Z}[\sqrt{2}]$.

Here I think I use the fact that $\mathbb{Z}[\sqrt{2}]$ is an euclidean ring.

Is anyone could help me at this point?

Answer 1

Since our polynomial has degree $2$, it is enough to show its zeros are not in $\mathbb{Z}[\sqrt{2}]$.

Suppose to the contrary that $2^{1/4}=a+b\cdot 2^{1/2}$, where $a$ and $b$ are integers. Then $$2=a^4+4a^3b\cdot2^{1/2}+12a^2b^2+8ab^3\cdot 2^{1/2}+4b^4.$$ By the irrationality of $\sqrt{2}$, we have $$2=a^4+12a^2b^2+4b^4.$$ This has no integer zeros. For clearly cannot have $a$ odd. And if $a$ is even then $4$ divides he right-hand side, but $4$ does not divide $2$.

Alternately, we can work with the equation $4a^3b+8ab^3=0$.

Answer 2

$x^2 - \sqrt{2}$ is an Eisenstein polynomial with respect to the prime $\sqrt{2}$.