Deducing whether elements are prime in $\mathbb{Z}[\sqrt{-2}]$

Question

Let $ \alpha_1 = 3 + 2\sqrt{-2}, \alpha_2 = 1 + 2\sqrt{-2}$, how do I deduce if these elements are prime in $\mathbb{Z}[\sqrt{-2}]$?

I've tried showing that they aren't irreducible, which in turn would show that they aren't prime, but I'm not making any headway.

Is there a standard way of tackling a problem like this?

Answer 1

Hint: Consider $N(\alpha_1)=3^2+2\cdot2^2=17$ and $N(\alpha_2)=1^2+2\cdot2^2=9$.

You now need to prove these properties for $N$:

• $N(\alpha \beta) = N(\alpha) N(\beta)$

• $N(\alpha) = 1$ iff $\alpha$ is a unit in $\mathbb{Z}[\sqrt{-2}]$

• If $N(\alpha)$ is a prime, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$

Therefore, $\alpha_1$ is irreducible, because $N(\alpha_1)=17$ is prime. Since $\mathbb{Z}[\sqrt{-2}]$ is a PID because it is Euclidean, being prime is equivalent to being irreducible.

To handle $\alpha_2$, we need to consider the elements that have norm $3$. These are exactly $\pm 1 \pm \sqrt{-2}$. We find that $ 1+ 2\sqrt{-2} = (-1 + \sqrt{-2})(1 - \sqrt{-2}) $ and so $\alpha_2$ is reducible and so not prime.