Abstract Algebra proof for quadratic extensions 
Let $n$ be a square free integer and alpha belong to $\mathbb{Z}[\sqrt{n}]$. If $N(\alpha)$ is a prime integer, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{n}]$. 

Here $N(\alpha)$ denotes the norm of $\alpha$.
How can I show this? I'll get back in the morning because it's super late right now! 
 A: The key idea is that multiplicative maps preserve multiplicative properties (a fact which is brought to the fore in divisor theory). Here we wish to pullback the property of being an atom (irreducible) along the multiplicative norm map $\rm\:N\:$ (recall that prime $\iff$ atom in a UFD, here $\rm\:Na\in \mathbb Z\:$).   
Recall, by definiton, an atom (irreducible) is a nonunit which can't be split into nontrivial factors, i.e. $\rm\ x = y\ z\ \Rightarrow\ y|1\:\ or\:\ z|1\:,\:$ i.e. $\rm\:y\:$ or $\rm\:z\:$ is a unit (or, equivalently, $\rm\:x\:|\:y\:\ or\:\ x\:|\:z).$ Now consider
HINT $\ $ atom $\rm Na,\:\ a = bc\ \Rightarrow\ Na = Nb\ Nc\ \Rightarrow Nb|1\ or\ Nc|1\ \Rightarrow\ b|1\ or\ c|1\ \Rightarrow\:$ atom $\rm\:a$
In fact much of the multiplicative structure of a number ring is reflected in its monoid of norms. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For references (Bumby and Dade, Lettl, Coykendall) see my sci.math post on 19 Dec 2007.
A: The norm is multiplicative. So, if $\alpha=\beta\gamma$ then $N(\alpha)=N(\beta) N(\gamma)$. If $N(\alpha)$ is prime, then one of $N(\beta)$ or $N(\gamma)$ is $\pm 1$. You now need that $N(\beta)=\pm 1$ iff $\beta$ is a unit, which comes directly from $N(\beta)=\beta\bar{\beta}$.
