Question about $p$-adic numbers and $p$-adic integers I've been trying to understand what $p$-adic numbers and $p$-adic integers are today. Can you tell me if I have it right? Thanks.
Let $p$ be a prime. Then we define the ring of $p$-adic integers to be 
$$ \mathbb Z_p = \{ \sum_{k=m}^\infty a_k p^k \mid m \in \mathbb Z, a_k \in \{0, \dots, p-1\} \} $$
That is, the $p$-adic integers are a bit like formal power series with the indeterminate $x$ replaced with $p$ and coefficients in $\mathbb Z / p \mathbb Z$. So for example, a $3$-adic integers could look like this: $1\cdot 1 + 2 \cdot 3 + 1 \cdot 9 = 16$ or $\frac{1}{9} + 1 $ and so on. Basically, we get all natural numbers, fractions of powers of $p$ and sums of those two.
This is a ring (just like formal power series). Now we want to turn it into a field. To this end we take the field of fractions with elements of the form 
$$ \frac{\sum_{k=m}^\infty a_k p^k}{\sum_{k=r}^\infty b_k p^k}$$
for $\sum_{k=r}^\infty b_k p^k \neq 0$. We denote this field by $\mathbb Q_p$. 
Now as it turns out, $\mathbb Q_p$ is the same as what we get if we take the ring of fractions of $\mathbb Z_p$ for the set $S=\{p^k \mid k \in \mathbb Z \}$. This I don't see. Because then this would mean that every number $$ \frac{\sum_{k=m}^\infty a_k p^k}{\sum_{k=r}^\infty b_k p^k}$$ can also be written as $$ \frac{\sum_{k=m}^\infty a_k p^k}{p^r}$$
and I somehow don't believe that. So where's my mistake? Thanks for your help.

I want to emphasize that $\mathbb Z_p$ is not just $\mathbb F_p[[X]]$ in disguise, though the two rings share many properties. For example, in the $3$-adics one has
\[
(2 \cdot 1) + (2 \cdot 1) = 1 \cdot 1 + 1 \cdot 3 \neq 1 \cdot 1.
\]
I know three ways of constructing $\mathbb Z_p$ and they're all pretty useful. It sounds like you might enjoy the following description:
\[
\mathbb Z_p = \mathbb Z[[X]]/(X - p).
\]
This makes it clear that you can add and multiply elements of $\mathbb Z_p$ just like power series with coefficients in $\mathbb Z$. The twist is that you can always exchange $pX^n$ for $X^{n + 1}$. This is the “carrying over” that Thomas mentions in his helpful series of comments.
