Hurkyl

For every nonzero x in [itex]\mathbb{Q}[\sqrt{-3}][/itex], there is exactly one congruence class modulo x. This is because x is invertible, and so it divides everything.

The set you describe is the ring of integers in [itex]\mathbb{Q}[\sqrt{-3}][/itex]. Equivalently, it is the ring [itex]\mathbb{Z}[ (1 + \sqrt{-3}) / 2][/itex].

The number field [itex]\mathbb{Q}[\sqrt{-3}][/itex] consists of all numbers of the form [itex]a + b \sqrt{-3}[/itex] where a and b are arbitrary rational numbers.


I don't know how well I can help; I really have no idea how this subject is presented if it doesn't assume abstract algebra as a prerequisite. Hopefully someone else can chime in.


Anyways, I think some of what I said is still applicable; you can try and just start at the definitions and grind your way to an answer. And it might help immensely if you can find a rational integer that's in the same congruence class of zero.

What is your definition of congruence class, btw?