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

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.

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.
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.