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I'm trying to get through a proof of Gauss' that certain primes can be written as the sum of two squares. An assumption is that

the order of $\mathbb{Z}[i]/(a+bi)$ is $a^2+b^2$.

I get that $(a+bi)(a-bi)=a^2+b^2$, so this places a bound on the order of integers with no imaginary part. But since $b$ isn't a unit, it doesn't seem like this finishes the proof.

Any hints?

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Do you know what the ideal $(a + bi)$ looks like in the complex plane? – Qiaochu Yuan Jul 20 '11 at 1:10
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Out of curiosity: how does this proof use the order of these quotients? The proof I'm thinking of doesn't need that at all. – Dylan Moreland Jul 20 '11 at 1:21
@Dylan: It is used to show that $\mathbb{Z}[i]$ is a Euclidean domain. – Xodarap Jul 20 '11 at 2:50
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@Xodarap Interesting. Here's what I had in mind: to divide $a$ by $b \neq 0$ I need $q, r \in \mathbf{Z}[i]$ such that $a = qb + r$ and $|r| < |b|$. Then it's enough to find a $q$ such that $|a/b - q| < 1$. But the farthest any point in the complex plane can be from the lattice $\mathbf{Z}[i]$ is $1/\sqrt{2}$. – Dylan Moreland Jul 20 '11 at 3:10

marked as duplicate by YACP, Dennis Gulko, Amzoti, Asaf Karagila, Ｊ. M. Apr 23 at 0:19

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3 Answers

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HINT $\ $ If $\rm\ (a,b) = 1\ $ then $\rm\ x^2 = -1,\ a = b\:x\ \iff\ a^2 = -b^2,\ x = a/b\:.\ $ Therefore

$$\rm\mathbb Z[i]/(a-b\:i)\ \cong\ \mathbb Z[x]/(x^2+1,\ a-b\:x)\ \cong\ \mathbb Z[x]/(a^2+b^2,x-a/b)\ \cong\ \mathbb Z/(a^2+b^2) $$

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I tried to prove your last isomorphism by the first isomorphism theorem, but I have difficulty showing surjectivity since I don't really know the elements of $\mathbb Z[x]/(a^2+b^2,x-a/b)$ - it doesn't seem like $\mathbb Z[x]/(x-a/b)\cong\mathbb Z$, for example. How can I attack this? – Xodarap Jul 20 '11 at 3:10
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HINT $\ $ It's simply $\rm\:\mathbb Z[x]/(n,x-m)\ \cong\ \mathbb Z/n\ $ since $\rm\ (a,b)=1\ \Rightarrow\ (a^2+b^2,b)\ =\ (a^2,b)\ =\ 1\ \ $ hence $\rm\ 1/b\ $ exists $\rm\:(mod\ a^2+b^2)\:.\:$ So put $\rm\ n = a^2+b^2,\ \ a/b\equiv m\ (mod\ n)\:.$ – Gone Jul 20 '11 at 3:17
It might be better to write this as two successive quotients, as $x - a/b$ isn't really an element of $\mathbf{Z}[x]$. Great answer! – Dylan Moreland Jul 20 '11 at 3:21
@Dylan $\rm\:a/b\:$ denotes an integer $\rm\:m\equiv a/b\ (mod\ n)\:$ so it is correct as written - see my prior comment. – Gone Jul 20 '11 at 3:26
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Hint: The Gaussian integers are a Euclidean domain with respect to the norm. Hence given $z\in \mathbb Z[i]$ there are $q$ and $r$ in $\mathbb Z[i]$ such that $z=q(a+bi)+r$, with $r=0$ or $N(r)<N(a+bi)$.

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I actually wanted a proof of this question so that I could prove $N(a+bi)=a^2+b^2$ made the domain Euclidean :-) – Xodarap Jul 20 '11 at 2:47
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Expanding on Qiaochu's comment, but trying to leave you some room to think:

$\mathbb{Z}[i]$ is a square lattice in the complex plane, and the ideal $(a+bi)$ is a square sublattice. (Why?) The index of $(a+bi)$ in $\mathbb{Z}[i]$ is the number of elements of $\mathbb{Z}[i]$ inside a fundamental square of $(a+bi)$, including part but not all of its boundary. (Why? And what's with the boundary funny business?) This is the same as the ratio of the area of the fundamental square of $\mathbb{Z}[i]$ to the area of the fundamental square of the sublattice $(a+bi)$. (Why?)

$\mathbb{Z}[i]$'s fundamental square is a unit square, so the question is really about the area of $(a+bi)$'s fundamental square. The line from $0$ to $a+bi$ is a side of this square...

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I recall you mentioning Artin in another question (math.stackexchange.com/questions/50757/…). Chapter 11 has a lot of stuff relevant to this question: section 2 has a diagram showing what $(2+i)$ looks like inside $\mathbb{Z}[i]$, as well as an expanded version of the proof that $\mathbb{Z}[i]$ is a Euclidean domain mentioned by Dylan Moreland; section 5 works through which primes are sums of squares; and section 10 relates the index of a sublattice to areas of fundamental regions as I have here (though he also leaves the proof as an exercise). – Ben Blum-Smith Jul 20 '11 at 15:39
Thanks Ben! Artin's geometric proof on pages 397-398 (similar to what you have here) is much easier to understand (for me) than the abstract nonsense argument. I'm still working through a couple things, but hopefully this will do it... – Xodarap Jul 20 '11 at 17:26
I had a question that was too long for a comment - could you take a look if you get a chance? – Xodarap Jul 21 '11 at 14:16

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