$\mathbb{Z}[2\sqrt2]$ is not a UFD.

Question

Dummit and Foote q9.3.1

Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $\text{Prove that if a(x)} \not \in \text{R[x] then R is not a U.F.D.}$ Deduce that $\mathbb{Z}[2\sqrt2]$ is not a U.F.D.

How to do the last part? 

$\mathbb{Z}[2\sqrt2]$ is integral domain. $F$ be its quotient field. To find $p(x)\in \mathbb{Z}[2\sqrt2][x](=\mathbb{Z}[2\sqrt2, x])$, $a(x) \not \in \mathbb{Z}[2\sqrt2, x]$, $a(x), b(x) \in F[x]$ of smaller degree than $p(x)$ s.t. $p(x) = a(x)b(x)$. $deg_x(p(x)) \geq 2$ because it can't be $0$ or $1$. 

Please give a hint how to find $p(x), a(x), b(x)$. Please do not give solution.

Thanks!

This is a direct solution which is maybe not expected here.

Answer 1

Hint: Think about $x^2+2\sqrt 2 x + 2$. Full solution:

Let $p(x)= x^2+2\sqrt 2x +2$
Then, $a(x)=b(x)=x+\sqrt 2 \not \in \mathbb Z[2\sqrt 2,x]$.