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 \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)

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- anonymous

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- 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.

- 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.

- KingGeorge

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

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## More answers

- 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:\[
\mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}[i]\;|\forall m\in\mathbb{Z}[i], n\equiv m\;(\!\!\!\!\!\!\mod z)\}
\]

- KingGeorge

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

- anonymous

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

- 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.

- KingGeorge

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

- KingGeorge

I must go, but I will think about this.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- KingGeorge

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

- anonymous

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

- anonymous

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

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