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Let $a+bi=α∈\mathbb{}[i]$ be a Gaussian integer. why is it that $N(α)=a^2+b^2$ is equal to the cardinality of $\mathbb{Z}[i]/(α)?$ My question can be generalized to quadratic integers in general, and I'm not sure to what extent it goes further. An answer addressing any of these points, or leading me to a good source, would be valuable.

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See my answer at math.stackexchange.com/questions/74406/… for an explanation that covers the case you talk about. (The observation that the norm of a Gaussian integer is positive is also relevant. Also see my comment there.) –  Barry Smith Apr 17 '12 at 2:03
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I think there is a geometric argument that works for the Gaussian case, though I don't recall it. –  jerome d Apr 17 '12 at 2:30
    
Would love to know a geometric argument if you could remember or point me to a source. –  Eric Gregor Apr 17 '12 at 18:53

3 Answers 3

up vote 7 down vote accepted

For an integer $n$, the quotient group ${\mathbf Z}[i]/(n)$ clearly has order $n^2$. Now consider the chain of ideals $({\rm N}(\alpha)) \subset (\alpha) \subset {\mathbf Z}[i]$. The index $[{\mathbf Z}[i]:{\rm N}(\alpha)]$ is $({\rm N}(\alpha))^2$. Through multiplication by $\alpha$, ${\mathbf Z}[i]/(\overline{\alpha}) \cong (\alpha)/({\rm N}(\alpha))$ as additive groups. Thus $$[{\mathbf Z}[i]:({\rm N}(\alpha))] = [{\mathbf Z}[i]:(\alpha)][(\alpha):({\rm N}(\alpha))] = [{\mathbf Z}[i]:(\alpha)][{\mathbf Z}[i]:(\overline{\alpha})].$$ Complex conjugation induces an isomorphism of ${\mathbf Z}[i]/(\alpha)$ with ${\mathbf Z}[i]/(\overline{\alpha})$ (as rings or, sufficient for our purposes, as additive groups). Thus the above displayed equation implies $$({\rm N}(\alpha))^2 = [{\mathbf Z}[i]:(\alpha)]^2.$$ Now take positive square roots of both sides and you're done.

This method applies without essential changes to show for any quadratic ring ${\mathcal O}$ (by which I mean a subring of rank 2 inside a quadratic field, not necessarily the full ring of integers) and nonzero $\alpha$ in ${\mathcal O}$ that $\#({\mathcal O}/(\alpha)) = |{\rm N}(\alpha)|$. In the proof replace complex conjugation with whatever the conjugation operation is on the quadratic ring. The absolute value is essential if you're in a real quadratic field, where the norm function can take negative values. For example, $\#({\mathbf Z}[\sqrt{2}]/(1+5\sqrt{2})) = |-49| = 49$. This approach does not tell you a set of representatives for the cosets mod $\alpha$, but you didn't ask for that.

In order to go beyond quadratic rings, say to integers in a number field, this method breaks down because it depends on the crutch of the conjugation operation. It's true that in any order ${\mathcal O}$ in a number field $K$, $\#({\mathcal O}/(\alpha)) = |{\rm N}_{K/\mathbf Q}(\alpha)|$, but the method above doesn't directly extend to that setting. You could first prove the result for a larger Galois extension containing $K$ and then use that to derive the case of interest, or you just need other techniques (e.g., structure theorem for torsion modules over a PID). The trick above for the quadratic case is nice since it gives you the result fairly painlessly (no need to find an explicit set of representatives, which can be tedious for ${\mathbf Z}[i]/(\alpha)$ if the real and imaginary parts of $\alpha$ are not relatively prime, and then would you have to start all over again if you wanted to work in ${\mathbf Z}[\sqrt{10}]\dots$).

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This is a great argument, thanks. If you know of the geometric one that jay wendt mentioned, that would be interesting to know as well. Also, I would be very interested to know what the set of representatives would be for the cosets mod $\alpha$! I could make a separate question, if you like. –  Eric Gregor Apr 17 '12 at 19:14
    
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A geometric argument is based on Pick's theorem. For nonzero Gaussian integer $\alpha$, coset reps for ${\mathbf Z}[i]/(\alpha)$ are lattice points in the parallelogram with two edges the vectors $\alpha$ and $i\alpha$. You also want to count all non-vertex lattice points on those two edges and their common vertex 0 just once (not twice). If you accept this is a complete set of coset reps, then $\#({\mathbf Z}[i]/(\alpha))= I+B/2-1$, where $I$ is the total number of interior lattice pts and $B$ is the total number of bdy lattice points. By Pick this is the area, and area = ${\rm N}(\alpha)$. –  KCd Apr 19 '12 at 2:40

It is worth emphasis that the standard proof in KCd's answer has a very suggestive presentation in "fractional" form, since the "cancellations" are analogous to cancellation of fractions, viz.

$$\rm\ \alpha\bar\alpha R\ \subset\ \alpha R\ \subset\ R $$

$$\rm \Rightarrow\ \ N(\alpha )^2 =\: \left|\frac{R}{\alpha\bar\alpha R}\right|\: =\: \left|\frac{R}{\alpha R}\right|\: \left|\frac{\alpha R}{\alpha\bar\alpha R}\right|\: =\: \left|\frac{R}{\alpha R}\right|\: \left|\frac{R}{\bar\alpha R}\right|\: =\: \left|\frac{R}{\alpha R}\right|^{\:\!2}$$

This is one reason for the use of quotient-like notation for quotient structures.

A more general method is to rewrite the ideal as a module in Hermite normal form, employing well-known algorithms, e.g. see H. Cohen's book (vol. 1) or Lemmermeyer's lecture notes.

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If $\gcd(a,b)=1$, then I believe that $\{0,1,2,\dots,N(\alpha)-1\}$ is a complete set of representatives for $\mathbb Z[i]/(\alpha)$; this would prove that $N(\alpha)$ equals the cardinality in question.

In general I think the same idea holds with the complete set of representatives $\{x+yi \colon 0 \le x < gN(\alpha),\, 0\le y<g\}$ where $g=gcd(a,b)$.

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