# Why does the ideal $(a+bi)$ have index $a^2+b^2$ in $\mathbb{Z}[i]$? [duplicate]

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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## marked as duplicate by YACP, Micah, Davide Giraudo, Amzoti, Asaf KaragilaApr 21 '13 at 18:04

Thanks 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
Sure, I'm also particularly curious about an algebraic proof. – yunone Oct 29 '11 at 12:02
This question has been asked before. – lhf Oct 29 '11 at 17:28
@lhf Would you be so kind and give a link instead of just saying it? – Phira Oct 30 '11 at 12:51
@Phira, math.stackexchange.com/questions/23358/… as mentioned in one of the answers. – lhf Oct 30 '11 at 13:21

In How $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$ 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 How $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$ 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 2:42

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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I'm glad to see my questions are linked to :) – Fredrik Meyer Oct 29 '11 at 22:46

$x+iy$ 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
@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 8:17
@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 9:13
...see, otherwise the use of Pick's theorem suits me fine for now. – yunone Oct 30 '11 at 9:13
@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 9:46

If you know a bit of theory, there's a chain of equalities:

• The index of the 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 a 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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