# Prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$.

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 $3$.

1. $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.
2. if $\alpha\mid\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)\mid N(\beta)$ in $\mathbb{Z}$.
3. $\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.
4. 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.

• possible duplicate of math.stackexchange.com/questions/86383/… Jan 18, 2015 at 21:15
• Not really a duplicate, since here OP supplies much of a proof and asks for comments. Jan 18, 2015 at 21:49

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.

Yes, you are on the right track. All your reasoning makes sense to me.

By the properties of the norm, associates have the same norm. So the only possible associates in your list are $1 + \sqrt{-5}$ and $1 - \sqrt{-5}$.

Now determine all units of $\mathbb{Z}[\sqrt{-5}]$ by finding all integer solutions of $N(a + b\sqrt{-5}) = 1$. (The list will be quite short.)

Then check every unit $u$ for $u (1 + \sqrt{-5}) = (1 - \sqrt{-5})$. You will see that this never happens and thus, the two elements are not associate.

There are only two units in $\mathbb{Z}[\sqrt{-5}]$: 1 and $-1$. To obtain the associate of a number, you multiply that number by a unit other than 1, and in this domain there is only one choice: $-1$. So, for example, the associate of 2 is $-2$, the associate of 3 is $-3$, the associate of $1 - \sqrt{-5}$ is $-1 + \sqrt{-5}$ and the associate of $1 + \sqrt{-5}$ is $-1 - \sqrt{-5}$.

You've already proven 2, 3, $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are irreducible. The arguments by norm were sufficient. However, in some other domains, like $\mathbb{Z}[\sqrt{10}]$, it becomes very important to watch out for associates.

I would like to close with some examples of "composite" numbers in $\mathbb{Z}[\sqrt{-5}]$:

• $2 \sqrt{-5}$
• $5 + 5 \sqrt{-5}$
• 21
• 29