$\sqrt {-6}$ is not prime in $\mathbb{Z}+\mathbb{Z}\sqrt {-6}$

Question

Suppose $\sqrt{-6}|(a+b\sqrt{-6})(c+d\sqrt{-6})$. I need to show that $\sqrt{-6}$ does not divide $(a+b\sqrt{-6})$ and does not divide $(c+d\sqrt{-6})$. I thought you might arrive at some contradiction by supposing $\sqrt{-6}(x+y\sqrt{-6})=(a+b\sqrt{-6})(c+d\sqrt{-6})$. Equating coefficients, $x=ad+bc$ and $y=bd$. I did the same for $\sqrt{-6}(x'+y'\sqrt{-6})=(a+b\sqrt{-6})$ and $\sqrt{-6}(x''+y''\sqrt{-6})=(c+d\sqrt{-6})$, but I'm not getting anywhere. I'm not really sure where to go

Answer 1

$6=2\cdot 3$ is divisible by $\sqrt{-6}$, but neither $2$ nor $3$ is divisible by $\sqrt{-6}$.

You only need one counter-example. It's not possible for all examples of $\sqrt{-6}\mid XY$ to have $\sqrt{-6}\not\mid X$ and $\sqrt{-6}\not\mid Y$, because you could just pick $X=\sqrt{-6}$.