## anonymous 3 years ago Here's a little brain teaser, that I feel some would enjoy: Given some Gaussian Integer $$z\in \mathbb{Z}[i]$$ find and prove a closed-form expression for the number of equivalence classes in $$\mathbb{Z}[i]/z\mathbb{Z}[i]$$

1. anonymous

It might seem easy to show, but it's not as easy as you might think to prove. I will say, though, there *is* a pretty neat proof.

2. KingGeorge

Could you give me a brief explanation of what $$\mathbb{Z}[i]/z\mathbb{Z}[i]$$ means? Specifically what it means to do the mod operation.

3. KingGeorge

$\mathbb{Z}[i]=\{a+bi\;|\;a,b\in\mathbb{Z}\} \quad\text{and}\quad i=\sqrt{-1}$

4. anonymous

Yeah, sorry, I was out. But, here: So, we have, like @KingGeorge said, $\mathbb{Z}[i]=\{a+bi\;|\;a,b\in \mathbb{Z}\}\text{ where }i^2=-1$While we have:$\mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}[i]\;|\forall m\in\mathbb{Z}[i], n\equiv m\;(\!\!\!\!\!\!\mod z)\}$

5. KingGeorge

So in the complex numbers, modular arithmetic takes place as follows correct?$a+bi\pmod{c+di}\equiv\left[ a\mod{c}+bi\mod{d}\right]$

6. anonymous

I don't quite understand your case... But in the Gaussian Integers, the modulus has the same definition as the normal integers: $a\equiv b\mod c \iff c\,|\,(b-a)\\ a, b, c \in \mathbb{Z}[i]$

7. anonymous

Wait, why'd my previous reply not make it? I forgot to add to the previous other post that the norm of the integer must be less than that of the modulus.

8. KingGeorge

That's what I thought, but just wanted to verify.

9. KingGeorge

10. anonymous

So your definition isn't quite correct; it should be:$\large\mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}i:\forall m\in\mathbb{Z}[i]~~\text{with}~~|m|<|z|,~n\equiv m\pmod{z}\}$

11. anonymous

Yes, it's not, it'd be easier to show using division algorithm... but, frankly, I'm too lazy to re-type it.

12. anonymous

And I'm too anally autistic to not correct it :)

13. anonymous

Haha, I'll re-type it later, I guess...

14. KingGeorge

Just fyi for people reading, how I thought the modular arithmetic worked at first, was not correct.

15. anonymous

Here's the better definition: $\mathbb{Z}[i]/z\mathbb{Z}[i]=\{r\in\mathbb{Z}[i]\;|\;\forall n\in\mathbb{Z}[i], q\in\mathbb{Z}[i], r=n-qz, \,N(r)<\frac{1}{2}N(z)\}$

16. anonymous

(Quintessentially, you have to remember that these are the sets of a single element (Gaussian Integer), which represents an equivalence class)