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-   -   Congruence Classes in Quadratic Integers (http://www.physicsforums.com/showthread.php?t=174599)

Frillth Jun20-07 02:14 PM

Congruence Classes in Quadratic Integers
 
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?

Hurkyl Jun20-07 03:29 PM

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)].

Frillth Jun20-07 06:25 PM

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?

Hurkyl Jun20-07 06:27 PM

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?

Frillth Jun28-07 03:15 AM

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?


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