Quadratic integer ring with universal side divisor? It seems that in every paper mentioning universal side divisors, they are defined very succinctly and with a bunch of symbols, so that I remain completely confused as to what they are and how to find a concrete specific example of such a divisor.
An answer to another question talks about polynomial rings, which I don't really understand. 

Can someone give me an example of a quadratic integer ring (like $\mathbb{Z}[i]$ or $\mathbb{Z}[\sqrt{2}]$) that has a universal side divisor and how to find that specific divisor?

 A: In complex quadratic fields there are only finitely many such USDs, since they contain at most 6 units. For example, $1+2i$ is a USD in $R = {\mathbb Z}[i]$ since the residue classes of $R/(1+2i)$ are represented by 0 and the units $\pm 1$, $\pm i$.
In ${\mathbb Z}[\sqrt{2}]$ there are probably infinitely many USDs since a prime element modulo which $\varepsilon = 1 + \sqrt{2}$ is a primitive root is such a divisor.
If you're looking for a less stupid name consider calling them elements of Euclidean length 1 and look into the papers of Motzkin for an explanation.
A: What has really helped me understand universal side divisors are these two examples in $\mathbb{Z}$: 2 and 3. The units of $\mathbb{Z}$ are $-1$ and 1. Obviously all even numbers are multiples of 2. As for odd numbers, you just add $-1$ or 1 (either one) and bam, you got an even number. To get a multiple of 3 from a number that's not already a multiple of 3 you add $-1$ or 1 (only one or the other will work).
In $\mathbb{Z}[i]$ we have two more units to keep in mind, $i$ and $-i$. We also have to look at norms. Every number with an even norm is divisible by $1 + i$ (which itself has 2 for a norm). If a number has an odd norm, all you have to do is add $i$, $-1$, $-i$ or 1 (any of these) to get a number with an even norm, which is therefore divisible by $1 + i$.
