Any "primitive" examples of a complex number with more than one distinct factorization in $\mathbb{Z}[\sqrt{-5}]$? Apologies in advance if this question is too basic for those who know and understand ideals.
We all know that $6$ has two distinct factorizations in $\mathbb{Z}[\sqrt{-5}]$. We can multiply $6$ by some number like $3 + \sqrt{-5}$ to get a number like $18 + 6 \sqrt{-5}$, which technically "inherits" two distinct factorizations from $6$.
But is there some number $a + b \sqrt{-5}$ (with $a, b \in \mathbb{Z}$, $a \neq b \neq 0$, $\gcd(a, b) = 1$) that has more than one distinct factorization in $\mathbb{Z}[\sqrt{-5}]$?
 A: I just want to fill in some details, some parts of the process of arriving at the answer which surely Ted and Lubin were aware of, at least subconsciously.
First of all, $\mathbb{Z}[\sqrt{-5}]$ has class number $2$. This means that it is not a unique factorization domain (which you already know), but also that the "failure" of unique factorization is not as "bad" as in domains with higher class numbers.
Thus if a number in this domain has two distinct factorizations, both factorizations have the same amount of irreducible factors.
In looking for the norms of numbers that might satisfy the desired requirements, we can therefore eliminate norms with an odd number of factors in $\mathbb{Z}$ (out go the primes). We can further narrow down the list of potential norms by limiting to norms corresponding to real integers (like $6$) that have more than one factorization in $\mathbb{Z}[\sqrt{-5}]$.
And then we further narrow down by looking at those norms that can be expressed as a product of smaller norms corresponding to real integers that have more than one factorization in $\mathbb{Z}[\sqrt{-5}]$. The first few such primitive numbers are $6, 9, 14, 21$. Then, since $126 = 6 \times 21 = 9 \times 14$, we just find the complex numbers with those norms and see if we can get their products to give us a number with the specified requirements: $$(1 + \sqrt{-5})(4 + \sqrt{-5}) = (-2 + \sqrt{-5})(3 - \sqrt{-5}) = -1 + 5 \sqrt{-5}.$$
A: My strategy is much like Ted’s, I think. First look for a prime of $\Bbb Z$ that is not principal in $\Bbb Z[\sqrt{-5}]$. Like $7=(7,3+\sqrt{-5})(7,3-\sqrt{-5})$. Aha! What about $14=2\cdot7=(3+\sqrt{-5})(3-\sqrt{-5})$?
But I want to take an opportunity to point out that $\Bbb Z[\sqrt{-6}]$ furnishes a much easier example of a quadratic imaginary ring without unique factorization: after all $6=2\cdot3=(\sqrt{-6})^2$.
A: What about factoring $\alpha = 1+5\sqrt{-5}$? Note that $N(\alpha) = 126 = 6\cdot 21 = 14\cdot 9$. (I confess I wrote a small Mathematica program to give me factorizations of norms, found one with at least 3 factors, and then looked for the corresponding elements of the ring.)
