Prove that the class number of $\mathbb{Z}[\zeta_3]$ is $1$

Solution 1:

Here's a nice geometric argument, which you can find essentially in Klein's lecture on Ideal Numbers (Lecture VIII in Lectures of Mathematics, by Felix Klein, AMS Chelsea Publishing, AMS, 2000).

Consider the lattice $\mathbb{Z}[\zeta_3]$ in $\mathbb{C}$ ($\zeta_3 = \frac{-1+\sqrt{-3}}{2}$). Given $a,b\in\mathbb{Z}[\zeta_3]$, $a\neq 0$. Consider the points in the lattice of the form $(q_1+q_2\zeta_3)a$, and use them to tessellate the complex plane. Then locate $b$, and the closest corner of one of the squares to $b$, $qa$. Then let $r=b-qa$.

Here is a picture which is slightly off, since my lattice here is $\mathbb{Z}[i]$, not $\mathbb{Z}[\zeta_3]$, but it should give you an idea; it's taken from the slides for a talk I gave some years ago. It's a bit hard to see, but the small dots are the lattice points; the big dots represent the $\mathbb{Z}[\zeta_3]$ multiples of $a$, with $a$ the first large dot to the right of the one labeled $0$ and above the horizontal line. (Well, as I said,t he small dots are really in the position of the lattice points of $\mathbb{Z}[i]$, but you get the idea, I hope). The three labeled dots above zero are (counterclockwise) $a$, $(1+\zeta_3)a$, and $\zeta_3 a$. The dot corresponding to $b$ is circled in red. The blue dot is $qa$, the blue arrow is the vector corresponding to $qa$, and the green arrow is the vector corresponding to $r = b-qa$. (The "1" at the very bottom of the picture is the page number, so ignore it...)

$\hskip 0.7in$

How big can $r$ be? The tessellation is by parallelograms whose sides have length $|a|$ and $|\zeta_3 a|=|a|$, so the furthest that $b$ can be from the corner it is closest to is $\frac{|a|\sqrt{2}}{2}$. That is, $|r|\leq\frac{|a|\sqrt{2}}{2}\lt|a|$. From this it follows that $0\leq N(r)\lt N(a)$.

Thus, for all $a,b\in\mathbb{Z}[\zeta_3]$, there exists $q,r\in\mathbb{Z}[\zeta_3]$ such that $b = qa+r$, and $0\leq N(r)\lt N(a)$. So $N$ is a Euclidean function on $\mathbb{Z}[\zeta_3]$, the latter is Euclidean, hence a PID, hence the class number is $1$.

The same geometric argument can be used to show that $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt{-2}]$ are Euclidean, but it breaks down when you get to $\mathbb{Z}[\sqrt{-5}]$, because the size of the rectangles now allows the distance from $b$ to a corner to be larger than the length of $a$ (good thing, too, since $\mathbb{Z}[\sqrt{-5}]$ is not a UFD).