Show that $\sqrt{-6}$ is irreducible in $\mathbb{Z}+\mathbb{Z}\sqrt{-6}$ Suppose not. Then there exists $\alpha,\beta\in\mathbb{Z}+\mathbb{Z}\sqrt{-6}$ such that $\sqrt{-6}=\alpha\beta\implies\alpha,\beta$ are not units. I'm not really sure where to go from here. Any advice?
 A: The concept of norm is extremely important, including the fact that it is multiplicative (meaning that $N(m) N(n) = N(mn)$). Learn it, love it, live it.
What really makes things easy in $\mathbb{Z}[\sqrt{-6}]$ (which is the same thing as $\mathbb{Z} + \mathbb{Z}\sqrt{-6}$, right?) is the fact that there are only two units: 1 and $-1$. How do we know those are the only units? A unit has norm 1 or $-1$. The norm of a number of the form $a + b \sqrt{-6}$ is $a^2 + 6b^2$, so the norm can't be negative. The possible norms in $\mathbb{Z}[\sqrt{-6}]$ are 0, 1, 4, 6, 7, 9, 10, etc. (see Sloane's A002481). Both 1 and $-1$ have a norm of 1, no number in this domain can have a norm of $-1$. If $a > 1$ or $b > 1$, then $N(a + b \sqrt{-6}) > 1$.
2 and 3 are potentially prime because no number in this domain has a norm of 2 or 3. But, as you already know from the other question you asked, $2 \times 3 = -(\sqrt{-6})^2 = 6$, which means 2 and 3 are irreducible but not prime in this domain.
But what about $\sqrt{-6}$? Its norm is 6. If it's reducible, we can find two numbers, neither of them 1, such that $N(\alpha) N(\beta) = 6$. So we're looking for a number with a norm of 2 and another with a norm of 3. But we just saw that there are no numbers in that domain with that norm. Therefore, $\sqrt{-6}$ is irreducible.
If you really want to be sure, you can try each product $(a - b \sqrt{-6})(a + b \sqrt{-6})$ for $0 \leq a \leq 3$ and $b$ likewise.
A: A start: Use the fact that in general $N(\alpha\beta)=N(\alpha)N(\beta)$. Here $N(\gamma)$ is the norm of $\gamma$. The norm of $x+y\sqrt{-6}$, where $x$ and $y$ are integers, is $x^2+6y^2$.
