<s>[INST] 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 
 [/INST] 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 $\pm(8+3\sqrt{7})^k.$
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