TITLE: Prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$.
QUESTION [14 upvotes]: So the Norm for an element $\alpha = a + b\sqrt{-5}$ in $\mathbb{Z}[\sqrt{-5}]$ is defined as $N(\alpha) = a^2 + 5b^2$ and so i argue by contradiction assume there exists $\alpha$ such that $N(\alpha) = 2$ and so $a^2+5b^2 = 2$ , however, since $b^2$ and $a^2$ are both positive integers then $b=0$ and $a=\sqrt{2}$ however $a$ must be an integer and so no such $\alpha$ exists, same goes for $3$.
I already proved that 

$N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.
if $\alpha\mid\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)\mid N(\beta)$ in $\mathbb{Z}$.
$\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.
Show that there are no elements in $\mathbb{Z}[\sqrt{-5}]$ with $N(\alpha)=2$ or $N(\alpha)=3$. (I proved it above)

Now I need to prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible.
So I also argue by contradiction, assume $1 + \sqrt{-5}$ is reducible then there must exists Non unit elements $\alpha,\beta \in \mathbb{Z}[\sqrt{-5}]$ such that $\alpha\beta = 1 + \sqrt{-5} $ and so $N(\alpha\beta) =N(\alpha)N(\beta)= N(1 + \sqrt{-5}) = 6$ but we already know that $N(\alpha) \neq 2$ or $3$ and so $N(\alpha) = 6$ and $N(\beta) = 1$ or vice verse , in any case this contradicts the fact that both $\alpha$ and $\beta$ are both non units.I just want to make sure i am on the right track here. And how can i prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are not associate to each other.

REPLY [5 votes]: I've seen these same arguments in quite a few books, it all looks pretty standard issue to me.
But what you seem to be missing is the motivation for all of this, the why do we care. That motivation is this famous fact: $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5}).$$ It's also true that $$6 = -2 \times -3 = (-1 - \sqrt{-5})(-1 + \sqrt{-5}).$$ We've got four factorizations here, but how many of them are distinct factorizations into irreducibles?
Well, $6$ has a norm of $36$. Per the familiar arguments you've already quoted, $2, 3, 1 - \sqrt{-5}, 1 + \sqrt{-5}$ are irreducible with norms of $4, 9, 6, 6$ respectively, and $36 = 4 \times 9 = 6 \times 6$.
I suppose you also need a good definition of associates: in a commutative ring, $x$ and $y$ are associates if and only if there is a unit $u$ such that $x = uy$ (see https://www.proofwiki.org/wiki/Definition:Associate/Commutative_Ring). In an imgainary quadratic ring like $\mathbb{Z}[\sqrt{-5}]$, if $u$ is a unit and $x$ and $y$ are associates, then $N(u) = 1$ and $N(x) = N(y)$.
Clearly $2$ is not an associate of $3$ nor of $1 - \sqrt{-5}$ nor of $1 + \sqrt{-5}$. It's possible that $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are associates of each other. But don't jump to the conclusion that $N(x) = N(y)$ automatically means $x$ and $y$ are associates, the definition explicitly requires a unit $u$ such that $x = uy$. There are only two units in $\mathbb{Z}[\sqrt{-5}]$, and those are $-1$ and $1$. As it turns out, $(-1)(1 - \sqrt{-5}) = -1 + \sqrt{-5}$, not $1 + \sqrt{-5}$, so this absence of a suitable unit means that $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are not associates of each other.
Although we've shown four factorizations of $6$ in $\mathbb{Z}[\sqrt{-5}]$, only two of these are distinct because the other two consist of associates of numbers in the first two.