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LolWolf

  • 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 \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)

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  1. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    \[\mathbb{Z}[i]=\{a+bi\;|\;a,b\in\mathbb{Z}\} \quad\text{and}\quad i=\sqrt{-1}\]

  4. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    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. LolWolf
    • 3 years ago
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    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. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    That's what I thought, but just wanted to verify.

  9. KingGeorge
    • 3 years ago
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    I must go, but I will think about this.

  10. Herp_Derp
    • 3 years ago
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    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. LolWolf
    • 3 years ago
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    Yes, it's not, it'd be easier to show using division algorithm... but, frankly, I'm too lazy to re-type it.

  12. Herp_Derp
    • 3 years ago
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    And I'm too anally autistic to not correct it :)

  13. LolWolf
    • 3 years ago
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    Haha, I'll re-type it later, I guess...

  14. KingGeorge
    • 3 years ago
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    Just fyi for people reading, how I thought the modular arithmetic worked at first, was not correct.

  15. LolWolf
    • 3 years ago
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    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. LolWolf
    • 3 years ago
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    (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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