Factorization in prime elements of $\mathbb{Z}\left[\sqrt{p}\right]$ for a prime number $p$

Question

I'm having troubles with the following problem:

Let $p$ be a prime number in $\mathbb{Z}$, and $\alpha\in\mathbb{Z}\left[\sqrt{p}\right]$ which is not a unit. Prove that $\alpha$ have a factorization in irreducible elements of $\mathbb{Z}\left[\sqrt{p}\right]$.

At the beginning I though that that ring was a Euclidean Domain, but that fails for $\mathbb{Z}\left[\sqrt{5}\right]$. So, I don't know where to start now.

Thanks in advance.

Answer 1

Hint: Consider the norm $N(a+b\sqrt{p})=|a^2-b^2p|$. Prove that if $\delta$ divides $\alpha$, then $N(\delta) \le N(\alpha)$. Then use induction on $N(\alpha)$.

Or, more sophisticatedly, argue that $\mathbb{Z}\left[\sqrt{p}\right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.

Answer 2

Suppose $\alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that $$\alpha=a_1a_2$$ If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so $$\alpha = a_3a_4a_2$$ In general there is an increasing chain of ideals $$(a_1)\subseteq (a_3)\subseteq \cdots$$ Since $\mathbb{Z}[\sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing, $$\alpha = a_1a_2$$ with $a_1$ irreducible. Thus you can iterate this to write $$\alpha = a_1\cdots a_n$$ with $a_1,\ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals $$(a_{n,1})\subseteq (a_{n+1,2})\subseteq \cdots$$ This must also stabilize, until you end up with a factorization into irreducible elements.