Teichmüller character

Before we define the Teichmüller character, we begin with a corollary of Hensel’s lemma.

Corollary.

Let $p$ be a prime number. The ring of $p$-adic integers (http://planetmath.org/PAdicIntegers) ${\mathrm{Z}}_p$ contains exactly $p\mathrm{-}\mathrm{1}$ distinct $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$th roots of unity. Furthermore, every $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$th root of unity is distinct modulo $p$.

Proof.

Notice that ${\mathbb{Q}}_p$, the $p$-adic rationals, is a field. Therefore $f(x)=x^{p-1}-1$ has at most $p-1$ roots in ${\mathbb{Q}}_p$ (see this entry (http://planetmath.org/APolynomialOfDegreeNOverAFieldHasAtMostNRoots)). Moreover, if we let $a\in \mathbb{Z}$ with $1\le a\le p-1$ then $f(a)=a^{p-1}-1\equiv 0modp$ by Fermat’s little theorem. Since $f^{\prime }(a)=(p-1)\cdot a^{p-2}$ is non-zero modulo $p$, the trivial case of Hensel’s lemma implies that there exist a root of $x^{p-1}-1$ in ${\mathbb{Z}}_p$ which is congruent to $a$ modulo $p$. Hence, there are at least $p-1$ roots in ${\mathbb{Z}}_p$, and we can conclude that there are exactly $p-1$ roots.∎

Definition.

The Teichmüller character is a homomorphism of multiplicative groups:

$$\omega :{𝔽}_p^{\times }\to {\mathbb{Z}}_p^{\times }$$

such that $\omega \mathit{}\mathrm{(}a\mathrm{)}$ is the unique $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$th root of unity in ${\mathrm{Z}}_p$ which is congruent to $a$ modulo $p$ (which exists by the corollary above). The map $\omega$ is sometimes called the Teichmüller lift of ${\mathrm{F}}_p$ to ${\mathrm{Z}}_p$ ($\mathrm{0}\mathrm{mod}p$ would lift to $\mathrm{0}\mathrm{\in }{\mathrm{Z}}_p$).

Remark.

Some authors define the Teichmüller character to be the homomorphism:

$$\widehat{\omega }:{\mathbb{Z}}_p^{\times }\to {\mathbb{Z}}_p^{\times }$$

defined by

$$\widehat{\omega }(z)=\underset{n\to \mathrm{\infty }}{lim}{z}^{p^n}.$$

Notice that for any $z\in {\mathbb{Z}}_p^{\times }$, $\widehat{\omega }(z)$ is a $(p-1)$th root of unity:

$${(\widehat{\omega }(z))}^p={\left(\underset{n\to \mathrm{\infty }}{lim}{z}^{p^n}\right)}^p=\underset{n\to \mathrm{\infty }}{lim}{z}^{p^{n+1}}=\widehat{\omega }(z).$$

Thus, the value $\widehat{\omega }(z)$ is the same than $\omega (zmodp)$.