Possible Irreducible but NOT prime!

Question

UFDs are integrally closed

Using the main result proved in the link I want to show that: $ A $={$a+b\sqrt {2}$ |$ a \in \mathbb Z$ & $ b$ is an even integer} is a subring of $\mathbb Z[ \sqrt {2} ]$ BUT $A$ is NOT a UFD .

My thought: I was trying to find out an element which is irreducible BUT not prime in $A$ . But couldn't figure out how to use this result! Please help!

Answer 1

Hint $\ $ You can use it like this. If $A$ were a UFD it would satisfy said Rational Root Test, so any root of $\ x^2 - 2\ $ in the fraction field of $A$ would be in $A$. But the root $\,x = \sqrt{2} = \dfrac{2\sqrt 2}2\not\in A$.

Answer 2

It's easier just to find an element that you can factor in more than one way. Note that the ring you're looking at is ${\mathbb Z}[2\sqrt{2}] = {\mathbb Z}[\sqrt{8}]$. This suggests looking at $8 = \sqrt{8}^2$ and $8 = 2^3$.

And when we're there anyway, it looks like both $\sqrt{8}$ and $2$ are irreducible elements that are not prime.