Find the congruence classes $\displaystyle (\bmod{(3+\sqrt{-3}})/2)$ in $\displaystyle \mathbb{Q}(\sqrt{-3})$.

I'm having trouble on this one. I don't know where to start.

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- Oct 1st 2008, 11:15 PMPn0yS0ld13rCongruences classes, Quadratic Field
Find the congruence classes $\displaystyle (\bmod{(3+\sqrt{-3}})/2)$ in $\displaystyle \mathbb{Q}(\sqrt{-3})$.

I'm having trouble on this one. I don't know where to start. - Oct 2nd 2008, 07:49 AMThePerfectHacker
Do you perhaps mean $\displaystyle \mathbb{Z}(\sqrt{-3})$?

Because $\displaystyle \mathbb{Q}(\sqrt{-3})$ is a__field__and so any two elements are divisible by any non-zero number. Which would imply the congruence class is the entire field itself. And in general, the congruence class mod any unit is always the entire ring itself. - Oct 2nd 2008, 07:11 PMPn0yS0ld13r
I don't think its a typo, but I copied it exactly how the problem is written. I'm not sure if it makes any difference, but I think the problem assumes that we're only dealing with quadratic integers in $\displaystyle \mathbb{Q}(\sqrt{-3})$.

My book gives the following definition for divisibility in $\displaystyle \mathbb{Q}(\sqrt{d})$:

If $\displaystyle \alpha$ and $\displaystyle \beta$ are integers in $\displaystyle \mathbb{Q}(\sqrt{d})$, with $\displaystyle \alpha \neq 0$, we say that $\displaystyle \alpha$ divides $\displaystyle \beta$ if there is an integer $\displaystyle \gamma$ in $\displaystyle \mathbb{Q}(\sqrt{d})$ such that $\displaystyle \beta = \alpha\gamma$. - Oct 4th 2008, 06:13 PMThePerfectHacker
- Oct 4th 2008, 06:40 PMPn0yS0ld13r
Yes it is, because if $\displaystyle d \equiv 1 \bmod{4}$, then the integers of $\displaystyle \mathbb{Q}(\sqrt{d})$ are those numbers of the form $\displaystyle \dfrac{a+b\sqrt{d}}{2}$ where $\displaystyle a,b \in \mathbb{Z}$ and a and b are both even or both odd.

So $\displaystyle \dfrac{3+\sqrt{-3}}{2}$ is a quadratic integer in $\displaystyle \mathbb{Q}(\sqrt{-3})$.