Method for determining irreducibles and factorising in $\mathbb Z[\sqrt{d}]$ I know that $\mathbb Z[\sqrt{7}]$ is a UFD, and I can write the equation $(2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7}) $. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles?
It seems difficult when dealing with $\mathbb Z [\sqrt{d}] $ with $d$ positive, since the norm map isn't as nice (it's not always positive).
Thanks 
 A: The norm is good enough for getting a good idea about the factors: $2 + \sqrt{7}$ has
prime norm, hence is irreducible, and the same goes for the other three factors. Thus 
the factors with equal norm must be associates, i.e. their quotients are units. You can find these by writing down the corresponding fractions and making the denominators rational.
A: An irreducible of $\mathbb{Z}[\sqrt{7}]$ must be a prime of $\mathbb{Z}$, or must lie above a prime of $\mathbb{Z}$.
Since $\mathbb{Z}[\sqrt{7}]$  is the splitting field of $x^2-7$, you want to determine how $x^2-7$ factors modulo $p$, for $p$ prime. In other words, for which primes is $7$ a square modulo $p$?
If $p=2$ or $p=7$, $x^2-7$ factors as a perfect square; so $2$ and $7$ ramify; that is, there is a unique irreducible (up to units) that divides $2$ and that divides $7$. It is not hard to find $3+\sqrt{7}$ as a divisor of $2$: $(3+\sqrt{7})(3-\sqrt{7}) = 2$. Note that $N(3+\sqrt{7}) = 2$ is a prime, so that means that $3+\sqrt{7}$ is necessarily irreducible. And of course, $\sqrt{7}$ divides $7$, and is irreducible.
For $p\neq2, 7$, we have: for $p\equiv 1\pmod{4}$, 
$$\left(\frac{7}{p}\right) = \left(\frac{p}{7}\right) = \left\{\begin{array}{ll}
1 & \text{if }p\equiv 1,2,4\pmod{7}\\
-1 & \text{if }p\equiv 3,5,6\pmod{7}.
\end{array}\right.$$
For $p\equiv 3\pmod{4}$, 
$$\left(\frac{7}{p}\right) = -\left(\frac{p}{7}\right) = \left\{\begin{array}{ll}
1 & \text{if }p \equiv 3,5,6\pmod{7}\\
-1 &\text{if }p\equiv 1,2,4\pmod{7}
\end{array}\right.$$
So a rational prime $p$ splits into a square (times a unit) if $p=2$ or $p=7$; factors into two distinct irreducibles in $\mathbb{Z}[\sqrt{7}]$ if and only if $p\equiv 1, 3, 9, 19, 25, 27\pmod{28}$; and remains irreducible if $p\equiv 5, 11, 13, 15, 17, 23$.
The units are precisely the elements of norm $\pm 1$; in fact, it cannot be $-1$, since that would require $a^2 -7b^2 =-1$, which has no solutions modulo $7$. You can find, through the "usual" methods, that every unit is of the form $\pm(8+3\sqrt{7})^k.$
