index of the group of cyclotomic units in the full unit group

Let $K_n=\mathbb{Q}({\zeta }_{p^n})$ where ${\zeta }_{p^n}$ is a primitive$p^n$th root of unity, let $h_n$ be the class number of $K_n$ andlet ${𝒪}_n={𝒪}_{K_n}$ be the ring of integers in$K_n$. Let $E_n={𝒪}_n^{\times }$ be the group of units in$K_n$. The cyclotomic units are a subgroup $C_n$ of $E_n$ whichsatisfy:

• The elements of $C_n$ are defined analytically.

• The subgroup $C_n$ is of finite index in $E_n$. Furthermore,the index is $h_n^{+}$: Let $E_n^{+}$ be the group of units in$K_n^{+}$ and let $C_n^{+}=C_n\cap E_n^{+}$. Then$[E_n^{+}:C_n^{+}]=h_n^{+}$. Moreover, it can be shown that$[E_n:C_n]=[E_n^{+}:C_n^{+}]$ because $E_n={\mu }_{p^n}{E}_n^{+}$ (this isexercise 8.5 in [1]).

• The subgroups $C_n$ behave “well” in towers. Moreprecisely, the norm of $C_{n+1}$ down to $K_n$ is $C_n$. Thisfollows from the fact that the norm of ${\zeta }_{p^{n+1}}$ down to$K_n$ is ${\zeta }_{p^n}$.

Definition 1.

Let $p$ be prime and let $n\mathrm{\ge }\mathrm{1}$. Let ${\zeta }_{p^n}$ be aprimitive $p^n$th root of unity.

1. The cyclotomic unit group $C_n^{+}\subset K_n^{+}=\mathbb{Q}{({\zeta }_{p^n})}^{+}$ is the group of units generated by$-1$ and the units

   $${\xi }_a={\zeta }_{p^n}^{(1-a)/2}\frac{1-{\zeta }_{p^n}^a}{1-{\zeta }_{p^n}}=\pm \frac{\sin (\pi a/p^n)}{\sin (\pi /p^n)}$$

   with and $\gcd (a,p)=1$.

2. The cyclotomic unit group $C_n\subset K_n=\mathbb{Q}({\zeta }_{p^n})$ is the group generated by ${\zeta }_{p^n}$and the cyclotomic units $C_n^{+}$ of $K_n^{+}$.

Remark 1.

Let ${\sigma }_a:{\zeta }_{p^n}\to {\zeta }_{p^n}^a$ be an element of$\mathrm{Gal}(K_n/\mathbb{Q})$. Then:

$${\xi }_a={\zeta }_{p^n}^{(1-a)/2}\frac{1-{\zeta }_{p^n}^a}{1-{\zeta }_{p^n}}=\frac{{({\zeta }_{p^n}^{-1/2}(1-{\zeta }_{p^n}))}^{{\sigma }_a}}{{\zeta }_{p^n}^{-1/2}(1-{\zeta }_{p^n})}.$$

Remark 2.

Let $g$ be a primitive root modulo $p^n$. Let$a\equiv g^{r}mod{p}^n$. Then one can rewrite ${\xi }_a$ as:

$${\xi }_a=\prod _{i=0}^{r-1}{\xi }_g^{{\sigma }_g^i}.$$

In particular ${\xi }_g$generates $C_n^{+}/\{\pm 1\}$ as a module over$\mathbb{Z}[\mathrm{Gal}(\mathbb{Q}{({\zeta }_{p^n})}^{+}/\mathbb{Q})]$.

Notice that in order to show that the index of $C_n$ in $K_n$ isfinite it suffices to show that the index of $C_n^{+}$ in $K_n^{+}$ isfinite. Indeed, let $[K_n:\mathbb{Q}]=2d$. Since $K_n$ is a totallyimaginary field and by Dirichlet’s unit theorem the free rank of$E_n$ is $r_1+r_2-1=d-1$. On the other hand, $[K_n^{+}:\mathbb{Q}]=d$ and$K_n^{+}$ is totally real, thus the free rank of $E_n^{+}$ is also$d-1$. Therefore the free rank of $E_n^{+}$ and $E_n$ are equal. Aswe claimed before, the index $[E_n^{+}:C_n^{+}]$ is rather interestingto us.

Theorem 1 ([1],Thm. 8.2).

Let $p$ be a prime and $n\mathrm{\ge }\mathrm{1}$. Let $h_n^{\mathrm{+}}$be the class number of $\mathrm{Q}\mathit{}{\mathrm{(}{\zeta }_{p^n}\mathrm{)}}^{\mathrm{+}}$. The cyclotomicunits $C_n^{\mathrm{+}}$ of $\mathrm{Q}\mathit{}{\mathrm{(}{\zeta }_{p^n}\mathrm{)}}^{\mathrm{+}}$ are a subgroup of finiteindex in the full unit group $E_n^{\mathrm{+}}$. Furthermore:

$$h_n^{+}=[E_n^{+}:C_n^{+}]=[E_n:C_n].$$

In the proof of the previous theorem one calculates the regulatorof the units ${\xi }_a$ in terms of values of Dirichlet L-functionswith even characters. In particular, one calculates:

$$R(\{{\xi }_a\})=\pm \prod _{\chi \ne {\chi }_0}\frac{1}{2}\tau (\chi )L(1,\overline{\chi })=h_n^{+}\cdot R^{+}$$

where in the last equality one uses the properties of Gausssums and the class number formula in terms of DirichletL-functions evaluated at $s=1$. This yields that $R(\{{\xi }_a\})$ innon-zero, therefore the index in $E_n^{+}$ is finite and moreover

$$h_n^{+}=\frac{R(\{{\xi }_a\})}{R^{+}}=[E_n^{+}:C_n^{+}]=[E_n:C_n].$$

An immediate consequence of this is that if $p$ divides $h_n^{+}$then there exists a cyclotomic unit $\gamma \in C_n^{+}$ such that$\gamma$ is a $p$th power in $E_n^{+}$ but not in $C_n^{+}$.