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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.

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    @ThomasAndrews Thanks a lot!!2012-07-25

2 Answers 2

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To define $\mathbb{Z}_p$ the summations should start at $k = 0$. In particular, it contains no negative powers of $p$.

As for your second question, it suffices to show that the inverse of a $p$-adic integer of the form $1 + a_1 p^1 + a_2 p^2 + ...$ is a $p$-adic integer. I'll write this as $1 - pz$ where $z$ is another $p$-adic integer. Then $\frac{1}{1 - pz} = 1 + pz + p^2 z^2 + p^3 z^3 + ...$

and this is a $p$-adic integer because only finitely many terms contribute to the coefficient of $p^k$ for any particular $k$. (I really am allowed to take this infinite sum because it converges $p$-adically.)

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    @Clark: the Euclidean norm cannot be extended to $\mathbb{Z}_p$.2012-07-25
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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.

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    I really like this because, when you then start talking about "inverse limits" definitions, you can also show that $\mathbb Z[[X]]$ is an inverse limit.2012-07-25