TITLE: Units of $\mathbb{Z}[\sqrt {-2}]$, and are $3,5$ irreducible in $\mathbb{Z}[\sqrt {-2}]$?
QUESTION [7 upvotes]: So for the first question I determined the units to be $+1$ and $-1$. I assigned $x = a + b \sqrt{-2}$ and proceeded to show $b = 0$ so $a^2 = 1$ or $-1$ so $x = 1$ or $-1$.
For the second part I am not sure what to do. Looking at the norm function $N$, we have $N(a + b\sqrt{-2}) = a^2 + 2b^2$ but I don't know how to go about with this information. I believe finding the norm would be important to determine the second part, right?

REPLY [3 votes]: Here's an approach that doesn't involve calculation of the norm.
For $3$ it's directly checkable that $3 = (1 + \sqrt{-2})(1 - \sqrt{-2})$.

To show that $5$ is irreducible, it is enough to show that the quotient ring $\mathbb Z[\sqrt{-2}]/(5)$ is a field.
We then write $\mathbb Z[\sqrt{-2}]$ as $\mathbb Z[X]/(X^2 + 2)$ and get $\mathbb Z[\sqrt{-2}]/(5) \simeq \mathbb Z[X]/(5, X^2 + 2) \simeq \mathbb F_5[X]/(X^2 + 2)$.
Therefore it suffices to show that the polynomial $X^2 + 2$ is irreducible in $\mathbb F_5[X]$.
Being a polynomial of degree $2$, one only needs to check that it doesn't have any root in $\mathbb F_5$.
This is however the fact that $-2$ is a quadratic non-residue mod $5$.

This approach has the advantage that it easily generalizes to other primes $p$ with $\left(\frac{-2}{p}\right) = -1$. The ultimate answer of course is given by reciprocity law (+ the fact that $\mathbb Z[\sqrt {-2}]$ is a Euclidean domain).