Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, Consider the set $\lambda P_n \subset \mathbb{Q}_p$, where $\lambda \in \mathbb{Q}_p^{\times}$ and $P_n$ is the set $\lbrace x \in \mathbb{Q}_p \mid \exists y \in \mathbb{Q}_p x= y^n\rbrace$. Is there a way to measure $\lambda P_n \cap B(r)$ for $r$ in $\mathbb{R}_{>0}$ the ball of radius $r$ (with Haar measure)?

Also, is $\mathbb{Q}_p^{\times}/ P_n^{\times}$ finite? what is its value? Thanks.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

What you are denoting $P_n$ is more usually denoted $\mathbf{Q}_p^{\times n}$, the image of the endomorphism $(\ )^n$ of $\mathbf{Q}_p^\times$.

To compute the quotient $\mathbf{Q}_p^\times/\mathbf{Q}_p^{\times n}$, recall that there is an exact sequence $$ 1\to\mathbf{Z}_p^\times\to\mathbf{Q}_p^\times\to\mathbf{Z}\to1 $$
which can be split by sending $1\in\mathbf{Z}$ to any uniformiser $\pi$ (such as $\pi=p$) of $\mathbf{Q}_p$. Further, there is an exact sequence $$ 1\to U_1\to\mathbf{Z}_p^\times\to\mathbf{F}_p^\times\to1 $$ which has a canonical splitting $\tau:\mathbf{F}_p^\times\to\mathbf{Z}_p^\times$ (the Teichmüller lift) and $U_r=1+p^r\mathbf{Z}_p$ is the group of units $\equiv1\pmod{p^r}$. Note that $U_1$ is a $\mathbf{Z}_p$-module which is free of rank $1$ if $p\neq2$; if $p=2$, then $U_1=\mu_2\times U_2$, where $\mu_2$ is the order-$2$ group consisting of $1$ and $-1$, and the $\mathbf{Z}_2$-module $U_2$ is free of rank $1$.

As a result, the choice of a uniformiser $\pi$ leads to an isomorphism $$ \mathbf{Q}_p^\times=U_1\times\mathbf{F}_p^\times\times\pi^{\mathbf{Z}} $$ which allows you to determine $\mathbf{Q}_p^\times/\mathbf{Q}_p^{\times n}$. For example, if $n$ is prime to $p$, then $$ \mathbf{Q}_p^\times/\mathbf{Q}_p^{\times n} =\mathbf{F}_p^\times/\mathbf{F}_p^{\times n}\times\pi^{\mathbf{Z}/n\mathbf{Z}} $$ and if $n=p$, then $$ \mathbf{Q}_p^\times/\mathbf{Q}_p^{\times p} =U_1/U_1^p\times\pi^{\mathbf{Z}/p\mathbf{Z}}\qquad p\neq2 $$ and $$ \mathbf{Q}_2^\times/\mathbf{Q}_2^{\times 2} =U_2/U_2^2\times\mu_2\times\pi^{\mathbf{Z}/2\mathbf{Z}}. $$ Note finally that $U_1^p=U_2$ if $p\neq2$ and $U_2^2=U_3$ if $p=2$. Similar considerations allow you to prove that $\mathbf{Q}_p^\times/\mathbf{Q}_p^{\times n}$ is finite for every $p$ and every $n>0$, and to compute its cardinality.

See for example Serre's Course in arithmetic, or any book on local fields.

share|improve this answer
    
Thank you for your nice answer. –  user16974 Jan 3 '12 at 11:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.