If $\,p\,$ is prime, is $\,p^n\mathbb{Z}_p=\mathbb{Z}_p\,$ for any positive integer $n$?

Question

Is $\,p^n\mathbb{Z}_p=\mathbb{Z}_p\,$ for any positive integer $n\,?$

$\mathbb{Z}_p =$ ring of $p$-adic integers, $\,p$ prime.

Thanks.

Answer 1

One way to see this is to consider $\mathbb{Q}_p$ as the completion of $\mathbb{Q}$ with respect to the $p$-adic absolute value $\lvert\cdot\rvert_p$. Then $\mathbb{Z}_p$ is defined as the unit ball inside $\mathbb{Q}_p$. And for $x\in p^n\mathbb{Z}_p$ write $$ x = p^nz. $$ Then $$ \lvert x\rvert_p = \lvert p^n\rvert_p\lvert z\rvert_p = \frac{1}{p^n}\lvert z\rvert_p\leq \frac{1}{p^n}. $$ However, you for example have that $\lvert 1\rvert_p = 1$.

You see that $p^n\mathbb{Z}_p$ "gets smaller and smaller" for larger $n$.

Answer 2

Think about it:

What must hold if $\;\;p^n\mathbb{Z}_p=\mathbb{Z}_p\quad?$

Is every element in $\mathbb Z_p$ also in $p^n\mathbb Z_p$? Consider, e.g., $1 \in \mathbb Z_p$. Is $1 \in p^n\mathbb Z_p$? If not, then it certainly cannot hold that $p^n\mathbb{Z}_p=\mathbb{Z}_p$.