Let $\zeta\_n = e^{i2\pi/n}$. What is the group of all units in the integral cyclotomic ring $\mathbb{Z}[\zeta\_n]$?


Here I like to know all the group elements for small $n$'s. For $n=1$ and $n=2$, the group is given by $\{1,-1\}$. For $n=3$, the group appears to be $\{\pm 1,\pm \zeta\_3, \pm \zeta\_3^2\}$. For $n=4$, the group is $\{\pm 1,\pm \zeta\_4\}$. What are the groups for larger $n$?



The group of all units of integral cyclotomic ring
As you suspect, identifying all the units in cyclotomic rings of order greater than $4$ is a non-trivial problem (I am ignoring order $6$ because, of course $\mathbb Z[\zeta\_6]$ is the same as $\mathbb Z[\zeta\_3]$, and similarly for all other orders twice an odd number.)


We can get just a taste of what is involved by considering order $5$, that is $\mathbb Z[\zeta\_5]$ = $\mathbb Z[\zeta\_{10}]$. Taking into account arithmetic inverses, we can identify a set of units given by $\zeta\_{10}^k$ where $\zeta\_{10}$ is any primitive tenth root of unity and $k$ is any integer. This gives a finite set of ten units.


But wait, there's more. If we add any complex conjugate pair, that is $\zeta\_{10}^k+\zeta\_{10}^{-k}$ for $k$ not a multiple of $5$, we will get a real sum having one of the forms


$\pm\tau,\pm(1/\tau)$


where $\tau=(1+\sqrt5)/2$ is the golden ratio (I am using an older notation here to avoid confusion with $\phi$ for the Euler totient function, which soon enters the discussion). And of course the presence of these reciprocal pairs implies that they are units, and so are their powers and products with the units derived from the tenth roots of unity. We are suddenly confronted with an infinite set all of whose elements must be units in $\mathbb Z[\zeta\_5]$:


$\color{blue}{\zeta\_{10}^k\tau^m}$


where $\zeta\_{10}, k, \tau$ are defined above and $m$ is another integer independent of $k$.


If we plot these points on a two-dimensional complex plane, we see that they are actually a subset of the tenfold symmetric quasilattice (the set of *all* elements in this ring would appear as the entire quasilattice). As such, they cannot all appear at unit distance from the origin when rendered onto the paper. We need the parent lattice having $\phi(5)=4$ dimensions to see geometrically that these elements all have equal norms.


Might there be more points than even these? Since $\phi(5)=4$, or in geometric terms four dimensions are sufficient to render the true unit norms, the set described in blue above (products formed from the primitive roots and one additional generator) turns out to be all the units in $\mathbb Z[\zeta\_5]$. Similarly we may construct four-dimensionally based sets using the roots of unity and one additional generator for $\mathbb Z[\zeta\_8]$ and $\mathbb Z[\zeta\_{12}]$ as $\phi(8)$ and $\phi(12)$ are also $4$ (again, $10$ is twice an odd number and thus $\mathbb Z[\zeta\_{10}]$ is not a distinct ring). For $\mathbb Z[\zeta\_{12}]$ the additional generator would actually be a complex number, as the real quasiperiodic ratio we might expect ($2+\sqrt3$) splits into two nonprimitive factors in the full cyclotomic field. We end with these:


$\mathbb Z[\zeta\_5]: \zeta\_{10}^k[(1+\sqrt5)/2]^m$


$\mathbb Z[\zeta\_8]: \zeta\_8^k(1+\sqrt2)^m$


$\mathbb Z[\zeta\_{12}]: \zeta\_{12}^k(1+\zeta\_{12})^m$


But these are the only cases which are that simple. With other orders having higher Euler totient functions, we need more complex (and geometrically higher-dimensional) structures to capture all the units.


