1. The problem statement, all variables and given/known data I need to find the congruence classes mod (3 + sqrt(-3))/2 in Q[sqrt(-3)]. 2. Relevant equations None known. 3. The attempt at a solution I'm not sure how to go about finding these congruence classes. I know that in the regular integers congruences classes mod x are just 0, 1, 2, ... x-1. Do I just try to find all of the quadratic integers with norm less than the norm of (3 + sqrt(-3))/2?
I assume you meant in Z[sqrt(-3)]? Or possibly in the ring of integers of Q[sqrt(-3)]? If nothing else, i think you can just grind out the answer straight from the definition: [tex]a \equiv b \mod{\frac{3 + \sqrt{-3}}{2}} \Longleftrightarrow \frac{3 + \sqrt{-3}}{2} \mid (a - b).[/tex] Or, equivalently, if (a - b) is in the ideal ((3 + sqrt(-3)) / 2). Actually, it might be easier to compute the ring Z[sqrt(-3)] modulo the ideal ((3 + sqrt(-3)) / 2). It's a relatively straightforward task if: (1) You've worked with finite fields (2) You write [itex]\mathbb{Z}[\sqrt{-3}] \cong \mathbb{Z}[x] / (x^2 + 3)[/itex] (3) You find the smallest (rational) integer in ((3 + sqrt(-3)) / 2). It might be doable even if you haven't done much with finite fields, and you simply find a single nonzero (rational) integer in that ideal. I think some people work better without using (2), but I think much better if I use (2). All of these ideas still apply if you are working with the ring of integers of Q(sqrt(-3)), rather than with the ring Z[sqrt(-3)].
My specific problem said that I need to work in the quadratic field Q[sqrt(-3)]. I know that this is different than Z[sqrt(-3)], because Z includes only integers of the form a + b*sqrt(-3) where a and b are integers, but Q[sqrt(-3)] includes integers of the from (a + b*sqrt(-3))/2, where a and b are both even or both odd. I've never heard or integer rings, finite fields, or ideals, and I have no idea what your statement 2 means, so I'm assuming I'm supposed to use a different method to find congruence classes. Any other ideas?
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?
This is mostly just intuition here, but I'm guessing that the congruence classes I'm looking for are: 0, 1, (1 + sqrt(-3))/2, and sqrt(-3). Now how do I go about testing whether these are correct?