existence and uniqueness of the reciprocal of a p-adic number

Question

Show that if $x\in Q_p$, then there exists $x^{-1}\in Q_p$.

My work: I want to show that there exists y such that $|xy|_p=1$(I'm not sure if this is equivalent to what question asks.)

Let $x=a_lp^{-l}+a_{-l+1}p^{-l+1}+...$by the definition of p-adic numbers. Since $|xy|_p=p^0$, then the lowest power of p of xy must be 0. So $xy=p^0\frac{r}{s}$ and $y=b_lp^l+b_{l+1}p^{l+1}...$

However, no matter what values of $(b_l,b_{l+1},...)$ are, $|xy|_p$ is always 1. So I'm wondering if the reciprocal of x is not unique or I got somewhere wrong when deciding $x^{-1}$.

Answer 1

$|xy|_p = 1$ doesn't mean $xy = 1$.

Assume $l=0$. Then $x =\sum_{n=0}^\infty a_n p^n= \lim_{N \to \infty} x_N$ is the $p$-adic limit of a sequence of integers, where $x_N =\sum_{n=0}^N a_n p^n$.

If $p\nmid a_0$ (so that $|x|_p = 1$) then its inverse $x^{-1}= y =\lim_{N \to \infty} y_N$ is the $p$-adic limit of any sequence of integers $(y_N)$ such that $|xy-1|_p = \lim_{N \to \infty} |x_Ny_N-1|_p = 0$.

Concretely the direct way is to take $y_N $ such that $p^N \mid x_Ny_N-1$ (ie. $y_N$ is an inverse of $x_N$ modulo $p^N$) obtaining $$|xy-1|_p = \lim_{N \to \infty} |x_Ny_N-1|_p = \lim_{N \to \infty} p^{-N} = 0$$

(note the fact $\lim_{N \to \infty} x_N $ converges in $\mathbb{Q}_p$ implies $\lim_{N \to \infty} y_N $ converges too)

Answer 2

Uniqueness: Assume $xy=xz=1$. Then $x(y-z)=0$ and $|x|_p|y-z|_p=0$, so either $x=0$ or $y-z=0$.

Existence: Let $x\in\Bbb Q_p$ with $0\ne x$. By definition, there exists a sequence $\{x_n\}_n$ with $x_n\in\Bbb Q$ and $x_n\to x$, i.e., $|x_n-x|_p\to 0$. Then for almost all $n$, $|x_n-x|_p<|x|_p$ and hence $|x_n|_p=|x|_p$. So we may assume wlog. that $|x_n|_p=|x|_p>0$ for all $n$. This allows us to define $y_n:=\frac1{x_n}$. Now $$|y_n-y_m|_p=\frac{|x_m-x_n|_p}{|x_nx_m|_p}=\frac1{|x|_p^2}\cdot |x_m-x_n|_p$$ and as, given $\epsilon>0$, there exists $N$ susch tat $|x_m-x_n|_p<|x|_p^2\epsilon$ for all $n,m>N$, we see that $|y_n-y_m|_p<\epsilon$ for $n,m>N$. That is, $\{y_n\}_n$ is Cauchy and so converges to some $y\in\Bbb Q_p$. As already $x_ny_n=1$ for all $n$, clearly $x_ny_n=1$.

Answer 3

My approach is quite different.

First, any nonzero element of $\Bbb Q_p$ is of form $p^ru$ where $|u|_p=1$, that is, $u$ is a $p$-adic unit. Thus it suffices to show that every $p$-adic unit has a reciprocal that’s also a $p$-adic unit.

If $u$ is a principal unit, i.e. of form $1+pw$ with $|w|\le1$, equivalently $u$’s canonical expansion starts with $1$, then you get the reciprocal of $u$ because $\frac1{1+pw}=1-pw+p^2w^2-p^3w^3+\cdots$, a $p$-adically convergent geometric series.

Finally, what if $u$ is a unit but not a principal unit, so that $u=m+pw$ for $m$ incongruent to both $0$ and $1$ modulo $p$? In this case, there is a $(p-1)$-th root of unity $\zeta$ with $\zeta\equiv u\equiv m\pmod p$. You can use Hensel’s Lemma to show this, or just let $m_0=m$ and $m_{i+1}=m_i^p$ for $i\ge0$ to get $\zeta=\lim_im_i$, since $\{m_i\}$ is a $p$-adically convergent sequence (not at all hard to show). Once you have $\zeta\equiv u\pmod p$, you also get $u/\zeta\equiv1\pmod p$, i.e. $u/\zeta$ is a principal unit, and has a reciprocal. Thus $u$ has a reciprocal.