Congruences classes, Quadratic Field

**Pn0yS0ld13r** · October 1st 2008, 11:15 PM

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.

**ThePerfectHacker** · October 2nd 2008, 07:49 AM

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.

**Pn0yS0ld13r** · October 2nd 2008, 07:11 PM

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 $\beta = \alpha\gamma$ .

**ThePerfectHacker** · October 4th 2008, 06:13 PM

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 $\mathbb{Q}[\sqrt{-3}]$

**Pn0yS0ld13r** · October 4th 2008, 06:40 PM

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

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