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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 closedform expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
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 closedform expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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
 3 years ago
Best ResponseYou've already chosen the best response.0Could 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\mathbb{Z}[i]=\{a+bi\;\;a,b\in\mathbb{Z}\} \quad\text{and}\quad i=\sqrt{1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0I don't quite understand your case... But in the Gaussian Integers, the modulus has the same definition as the normal integers: \[ a\equiv b\mod c \iff c\,\,(ba)\\ a, b, c \in \mathbb{Z}[i] \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, 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
 3 years ago
Best ResponseYou've already chosen the best response.0That's what I thought, but just wanted to verify.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0I must go, but I will think about this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, it's not, it'd be easier to show using division algorithm... but, frankly, I'm too lazy to retype it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And I'm too anally autistic to not correct it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haha, I'll retype it later, I guess...

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Just fyi for people reading, how I thought the modular arithmetic worked at first, was not correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's the better definition: \[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{r\in\mathbb{Z}[i]\;\;\forall n\in\mathbb{Z}[i], q\in\mathbb{Z}[i], r=nqz, \,N(r)<\frac{1}{2}N(z)\} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(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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