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P: 16,101

## Congruence Classes in Quadratic Integers

For every nonzero x in $\mathbb{Q}[\sqrt{-3}]$, 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 $\mathbb{Q}[\sqrt{-3}]$. Equivalently, it is the ring $\mathbb{Z}[ (1 + \sqrt{-3}) / 2]$.

The number field $\mathbb{Q}[\sqrt{-3}]$ consists of all numbers of the form $a + b \sqrt{-3}$ 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?