Prove that $a \equiv b \mod{p^j} \iff |a-b|_{p}\leq p^{-j}$

Question

Prove that $a \equiv b \mod{p^j} \iff |a-b|_{p}\leq p^{-j}$

Typically, when dealing with a congruence I go to the division statement. i.e $$a\equiv b\mod{p^j}\Rightarrow p^j|a-b \;\;\;(\star)$$ Moreover, I know that \begin{align} |a-b|_{p}&=p^{-vp(a-b)}\\ &\leq\max{(p^{-vp(a)},p^{-vp(a)})}\;\;\; (\star\star) \end{align}

However, I am having trouble with the key part that will connect $(\star)$ and $(\star\star)$.

I am hoping someone can help me fill in the missing parts. Thank you.

Answer 1

Instead of thinking of $a-b$ as a difference of two numbers, think of it as a single $p$-adic number. Then $a-b \equiv 0 \pmod {p^j}$ is exactly the statement that $p^j \mid (a-b)$, which is exactly the statement that the $p$-adic norm of $a-b$ is at least $j$ (and is more than $j$ if and only if additional factors of $p$ divide $a-b$, i.e. $a -b \equiv 0 \pmod{p^{j + \ell}}$).

Answer 2

it is like definition.

$p^j \mid a-b$ implies $a-b=p^tm$ where $t \geq j$ and $(m,p)=1$. Now $|a-b|_p=p^{-t}\leq p^{-j}$.