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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!

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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.

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I just wanted to let you know that your answer is probably over the OP's head...but I liked it very much and was thinking along similar lines. +1. :) – Pete L. Clark May 13 '11 at 16:45
@Pete Ha, tables turned, touche! Here I think that the hint is probably not over the OPs head - only the cultural background remarks - which can be ignored (but which may plant intuitive germs of more conceptual ideas that will later come to the fore). – Bill Dubuque May 13 '11 at 16:56

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 $N(\beta)=\beta\bar{\beta}$.

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