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Background

Definition: The function $N:\mathbb{Z}[\sqrt d] \to \mathbb{Z}$ given by

$$N(a+b\sqrt d)=(a+b\sqrt d)(a-b\sqrt d)=a^2 - db^2$$

is called the norm.

Exercise 39: Let $I$ be a nonzero ideal in $\mathbb{Z}[i]$. Show that the quotient ring $\mathbb{Z}[i]/I$ is finite.

Proof: Let $I$ be a nonzero principal ideal in $\mathbb{Z}[i],$ and let $a \in I.$ By the division algorithm, every element $b$ of $\mathbb{Z}[i]$ can be written as $b=ay+p,$ where $N(p)<N(a).$ Hence every element of $\mathbb{Z}[i]$ contains a $p$ with $N(p)<N(a),$ so it suffices to show that there are finitely many such $p.$ But this is clear: if $=w+iz$ has norm less than $n,$ then $w^2 + z^2<n$ and hence $|w|$ and $|z|$ are less than $n,$ hence there are certainly less that $4n^2 + 4n +1$ possibilities for $p$.

Questions:

For the proof above, I don't understand how the author got the number $4n^2 + 4n +1.$ Is the author counting the number of possibilities for $a, b$ in $a^2 - db^2?$ If so, how?

Thank you in advance

user26857
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Seth
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  • You do understand that this definition of the norm is for the ring $\Bbb{Z}[\sqrt{d}]$. And you have $d=-1$. – Jyrki Lahtonen Jan 18 '24 at 08:14
  • @JyrkiLahtonen I thought it is for any values of $d$, negative or positive. – Seth Jan 18 '24 at 09:02
  • Yes, the definition of the norm applies to all $d$, but your question is only about $d=-1$, so I felt it was a bit strange. Particularly because you wrote $N$ to be a function from $\Bbb{Z}[i]$ to $\Bbb{Z}$, – Jyrki Lahtonen Jan 18 '24 at 11:13
  • @JyrkiLahtonen I verbatium copied the definition from the textbook. I think the author did broke down the cases when $d>1,d \leq -1.$ – Seth Jan 18 '24 at 12:07

1 Answers1

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There are $2n+1$ possible values of $w$ that satisfy $|w| < n$, and likewise of $z$. These are the integers $\{-n, \ldots, 0, \ldots, n\}$, yielding a crude bound of $(2n+1)(2n+1) = 4n^2+4n+1$ values with norm less than $n$.

user26857
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Orange Mushroom
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  • oh so i was right that we should compute the different possibilités for $w,z$. Is just the number the author gave look so familiar, like I have seen it somewhere before. – Seth Jan 18 '24 at 09:05
  • $|w| < n \iff w\in{-(n-1), \ldots, 0, \ldots, n-1}$ and so the bound should be $(2n-1)^2$. – lhf Jan 19 '24 at 09:59
  • @lhf thank you for your comment. Sorry for the late reply. I just saw this now. I was going to create another post about the bounds. If $|w|< n,$ and $w=a+bi,$ and $|w|=a^2+b^2<n,$ should either the values of $a$ or $b$ each beless than or equal to $\frac{n-1}{2}?$ Or $w\in{-(n-1), \ldots, 0, \ldots, n-1}$ denotes the choices either of $a$ or $b$ can choose from? – Seth Jan 28 '24 at 20:12