Abstract Algebra – Prove $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are Irreducible in $\mathbb{Z}[\sqrt{-5}]$

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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

  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.

Best Answer

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

On your question about the associates

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.

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