# Thread: Congruences classes, Quadratic Field

1. ## Congruences classes, Quadratic Field

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

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

2. Originally Posted by Pn0yS0ld13r
Find the congruence classes $(\bmod{(3+\sqrt{-3}})/2)$ in $\mathbb{Q}(\sqrt{-3})$.

I'm having trouble on this one. I don't know where to start.
Do you perhaps mean $\mathbb{Z}(\sqrt{-3})$?

Because $\mathbb{Q}(\sqrt{-3})$ 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.

3. Originally Posted by ThePerfectHacker
Do you perhaps mean $\mathbb{Z}(\sqrt{-3})$?

Because $\mathbb{Q}(\sqrt{-3})$ 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.
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 $\mathbb{Q}(\sqrt{-3})$.

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

If $\alpha$ and $\beta$ are integers in $\mathbb{Q}(\sqrt{d})$, with $\alpha \neq 0$, we say that $\alpha$ divides $\beta$ if there is an integer $\gamma$ in $\mathbb{Q}(\sqrt{d})$ such that $\beta = \alpha\gamma$.

4. Originally Posted by Pn0yS0ld13r
My book gives the following definition for divisibility in $\mathbb{Q}(\sqrt{d})$:

If $\alpha$ and $\beta$ are integers in $\mathbb{Q}(\sqrt{d})$, with $\alpha \neq 0$, we say that $\alpha$ divides $\beta$ if there is an integer $\gamma$ in $\mathbb{Q}(\sqrt{d})$ such that $\beta = \alpha\gamma$.
But $\frac{3}{2} + \frac{\sqrt{3}i}{2}$ is not an integer in $\mathbb{Q}[\sqrt{-3}]$

5. Originally Posted by ThePerfectHacker
But $\frac{3}{2} + \frac{\sqrt{3}i}{2}$ is not an integer in $\mathbb{Q}[\sqrt{-3}]$
Yes it is, because if $d \equiv 1 \bmod{4}$, then the integers of $\mathbb{Q}(\sqrt{d})$ are those numbers of the form $\dfrac{a+b\sqrt{d}}{2}$ where $a,b \in \mathbb{Z}$ and a and b are both even or both odd.

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