In the comments on the question Why does this module structure have $352512$ elements?, it is mentioned that the index of the ideal generated by $a+bi$ in $\mathbb{Z}[i]$ has order $a^2+b^2$.
Is there a nice rigorous explanation of why this is so?
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1Thanks for asking, I would not mind to see an algebraic proof, too. I didn't realize that this was exactly Pick's formula until now, so I wanted it documented for myself. – Phira Oct 29 '11 at 12:00
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Sure, I'm also particularly curious about an algebraic proof. – yunone Oct 29 '11 at 12:02
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This question has been asked before. – lhf Oct 29 '11 at 17:28
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1@lhf Would you be so kind and give a link instead of just saying it? – Phira Oct 30 '11 at 12:51
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@Phira, http://math.stackexchange.com/questions/23358/how-mathbbzi-3-i-cong-mathbbz-10-mathbbz/23379#23379 as mentioned in one of the answers. – lhf Oct 30 '11 at 13:21
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To me it seems that that answer states $\mathbb{Z}[i]/(a - i b)$ has order $a^2+b^2$, but doesn't have an explicit proof of why. – yunone Oct 30 '11 at 13:25
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@lhf: I mentioned in my answer what exactly was proved there and what was not proved there. It is not exactly the same question. – Phira Oct 30 '11 at 13:28
4 Answers
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In Quotient ring of Gaussian integers it was shown that for $a,b$ coprime the quotient is actually isomorphic to $\mathbb Z / (a^2+b^2)$, but in general, it is still true that they have the same size.
If you look at the picture 
taken from the answer Quotient ring of Gaussian integers by quanta in the above-mentioned thread, you see that we want to count the lattice points in the square spanned by $a+bi$ and $-b+ai$.
As the points on the border have to be partially identified, it turns out we want to count interior points plus half the border points -1. (Since opposite sides of the square are identified, we want to count only half the border points, but we want to count only 1 of the 4 corners, so we have to subtract one.)
This gives exactly the area $\left(\sqrt{a^2+b^2}\right)^2=a^2+b^2$ of the square by Pick's theorem.
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This is really a nice way to look at it geometrically. Thanks for introducing me to Pick's theorem. – yunone Oct 30 '11 at 02:42
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One have $(a+ib)\mathbb Z[i]= (a+ib)\mathbb Z \oplus (-b+ia)\mathbb Z$, so the ideal $(a+ib)$ is the free $\mathbb Z$-submodule of $\mathbb Z[i] =\mathbb Z \oplus i\mathbb Z$ generated by $a+ib$ and $-b+ia$. Some basic algebra theory tells you that the index of this submodule is $\det \begin{pmatrix} a&-b\\b&a \end{pmatrix} = a^2+b^2$.
Edit : See here for the proof : Why is the determinant equal to the index?
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If you know a bit of theory, there's a chain of equalities:
- The index of the ideal is equal to the number of elements in the quotient ring
- The number of elements in the quotient ring is equal to the norm (over $\mathbb{Q})$ of the ideal
- The norm of an ideal is equal to the (ideal generated by the) norm of a generator
- The norm on $\mathbb{Q}(i)$ over $\mathbb{Q}$ is given by $\mathcal{N}(x + \mathbb{i} y) = x^2 + y^2$.
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$x+iy$ is part of the ideal generated by a+ib iff can be written as $(ac-bd)+i(cb+ad) = (a+ib)(c+id)$.
So $\begin{align} x&= ac-bd\\ y&= bc+ad \end{align} $
Solving for $c$ and $d$ we find
$\begin{align}
c&= (x+bd)/a\ d&=(ay-bx)/(a^2 + b^2) \end{align}$
Since $d$ needs to be an integer we have that
$a^2 + b^2 | ay-bx$
From Bézout's Identity we know that $ay-bx$ can be any integer if $a$ and $b$ are coprime, therefore we have $a^2 + b^2$ equivalence classes.
If $a$ and $b$ are not coprime (let $d$ be the GCD) we only have $(a^2 + b^2)/d$ equivalence classes.
I never used a math text editor so I'm sorry about the notation. I feel weird about not using the fact that $a|(x+bd)$ but the rest should be ok. I would be glad if someone can explain me how to write math here. Thanks.
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Are the details where $a$ and $b$ are coprime needed here? This condition doesn't seem to be mentioned elsewhere, and the result of the previous question I linked to seems to be wrong if there are only $(a^2+b^2)/d$ equivalence classes when $a$ and $b$ are not coprime. The module then only has $14688$ elements, and not $352512$. – yunone Oct 29 '11 at 23:24
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@yunone: Something went wrong with the conclusion here. Say, if $a=2$ and $b=0$ the index is surely four, not one. In that case the ideal consists of the gaussian integers with both even and real parts even, and that has 4 cosets. It may be a good idea to first consider the coprime case, and then deal with the common factor separately, or... – Jyrki Lahtonen Oct 30 '11 at 08:17
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@Jyrki I'm sorry, but I'm not sure what you're getting at. The only thing I was able to deduce was that $x+yi\equiv u+vi$ in $(a+bi)\mathbb{Z}[i]$ iff $a(x-u)+b(y-v)$ and $a(y-v)-b(x-u)$ are both $0\pmod{a^2+b^2}$. Matt E mentioned to me that it follows from the Euclidean Algorithm for $\mathbb{Z}[i]$, which I first understand that any coset has a representative whose norm is less than $a^2+b^2$ and there are $a^2+b^2$ nonnegative integers less than $a^2+b^2$, but this confused me since no representatve has norm $4k+3$, for instance. If you care to explain further, I'd be most gracious to... – yunone Oct 30 '11 at 09:13
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1@yunone: I agree that Pick's theorem (or Euclid's algorithm, or perhaps the argument by user10676) is best here. I was just chiming in that this answer is wrong. What I was suggesting is the following: If $gcd(a,b)=d$, then we can first draw the trivial conclusion that the ideal $I$ generated by $a+bi$ is an index $d^2$ subgroup in the ideal $J$ generated by $(a/d)+(b/d)i$. Then $gcd(a/d,b/d)=1$, and the arguments based on $a,b$ being coprime shows that ideal $J$ is of index $(a/d)^2+(b/d)^2$ in the ring $\mathbf{Z}[i]$. Putting these two pieces together gives the desired conclusion again. – Jyrki Lahtonen Oct 30 '11 at 09:46
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@Jyrki Thanks for that!, and for confirming that the answer isn't wholly correct. I notice you mention Euclid's algorithm, but how does it follow so easily from that? For say $\mathbb{Z}/m\mathbb{Z}$, it's clear this has size $m$ since there are only $m$ possible remainders modulo $m$. How does it work analogously in $\mathbb{Z}[i]$? After dividing some $x+yi$ by $a+bi$, I should get a remainder with norm less than $a^2+b^2$, but how would I know there are $a^2+b^2$ possible remainders? – yunone Oct 30 '11 at 09:56
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@yunone: Probably the best way to see that is to look at the figure in Phira's answer. The length of the diagonal of the square is $d=\sqrt{2(a^2+b^2)}$. And you also see that any point on the plane (in particular any point on the grid) is at a distance at most $d/2$ from one of the corners (=red dots). So if you need to "divide" a gaussian integer $w=u+iv$ by $a+bi$, then you locate the red dot closest to $w$. That is of the form $q(a+bi)$ for some gaussian integer $q$. Let $r$ be the separation vector. Then you have $w=q(a+bi)+r$, and $|r|\le d/2=\sqrt{(a^2+b^2)/2}$. – Jyrki Lahtonen Oct 30 '11 at 10:12
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