Prove $x^2-\sqrt2$ is irreducible over $\mathbb{Z}[\sqrt 2]$

Question

Dummit and Foote Q9.4.9

Prove that the polynomial $x^2-\sqrt2$ is irreducible over $\mathbb{Z}[\sqrt 2]$. (you may use the fact that $\mathbb{Z}[\sqrt 2]$ is a U.F.D.)

In a UFD 

1. every element can be written as a product of irreducibles

2. this decomposition is unique.

Do I use 1)? But I do not know the irreducibles in $\mathbb{Z}[\sqrt 2]$. 

Or do I use 2)? or something else?

I tried assuming factors of degrees 0 and 2, and of degrees 1 and 1. And solve for their coefficients. But since the factors may not be monic, the equations are complicated.

Please give a hint.

Thanks!

Answer 1

Something to help you get started:

An irreducible polynomial is one that cannot be factored into the product of two non-constant polynomials. To show this is irreducible, you would write $x^2-\sqrt{2}=(ax+b)(cx+d)$ for some $a,b,c,d\in\mathbb{Z}[\sqrt{2}]$ and hope to arrive at a contradiction.

Two polynomials are equal if and only if the coefficients of each $x^i$ are equal. So we would need $1=ac$, $0=ad+bc$, and $-\sqrt{2}=bd$.

The interesting equation there is $1=ac$. What values in $\mathbb{Z}[\sqrt{2}]$ have a multiplicative inverse, i.e. what are the options for $a$ and $c$?