Representation of p-adic integers

Question

Let $U_p:=\{(a_1,a_2,\dots) \in \Bbb Z_p:a_1 \neq 0\}$ and $x \in \Bbb Z_p \setminus \{0\}$. I don't understand why we can represent $x$ in an unique way as $x=p^{v_p(x)}u$, with $u \in U_p$ and $v_p(x) \in \Bbb N_0$. Can someone helps me please? Thanks a lot!

Answer 1

Suppose $x = (a_1 , a_2 \cdots )$ with $a_1 \neq 0$, then $x$ is already in $U_p$. It is expressed as $x=p^0 u$ with $u=x$.

Suppose $x = (0,a_2,a_3 \cdots )$ with $a_2 \neq 0$, then take $u=(a_2,a_3 \cdots)$. $p^1 u$ shifts everything one space to the right so $x=p^1 u$.

Continue on as $x=(0, \cdots 0, a_k , \cdots )$ with $a_k \neq 0$ being the first nonzero p-adic digit. Then $u=(a_k,a_{k+1},\cdots)$ and $x=p^{k-1} u$.

Now depends on how you are starting with $\nu_p (x)$ to see that this is the same thing.

Edit:

$U_p$ is given as a subset of $\mathbb{Z}_p$. You can think of it as the sphere and the rest of $\mathbb{Z}_p$ is the ball it encloses (which includes the sphere itself). So what this is doing is saying if you have a nonzero point in the ball, you can scale it up and get to the sphere.

Example:

$x= 0 + 0 p^1 + 0 p^2 + 1 p^3 + 1 p^4 + 0 p^5 + 1 p^6 + 0 p^7 \cdots 0 p^n \cdots$. In your notation this is $(0,0,0,1,1,0,1,0,0 \cdots )$ where everything else is 0. If you factor out $p^3$ you see $x= p^3 (1 + 1 p^1 + 0 p^2 + 1 p^3 + 0 \cdots )$. Or in the other notation $x=p^3 (1,1,0,1,0 \cdots)$. So $u=(1,1,0,1,0 \cdots)$ and $k=4$ because the first nonzero place was the 4th term.