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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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The $p$-adic integers have such form with $m=0$. $m<0$ are the general $p$-adic numbers. In particular, $\frac{1}{9}$ is not a $3$-adic integer. –  Thomas Andrews Jul 25 '12 at 15:06
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"I somehow don't believe that" doesn't constitute an attempt at a proof, so it's not clear to me in what sense there is a mistake to find. –  Qiaochu Yuan Jul 25 '12 at 15:10
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@ClarkKent No, that definition of addition does not work. Addition in $p$-adics has to be defined so that it is the same as addition when both $p$-adics are integers (aka where all but finitely many $a_k$ are zero.) –  Thomas Andrews Jul 25 '12 at 15:37
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@ClarkKent Yes, but your definition doesn't agree. If you add $4= 1+3$ and $5=2+3$ in $3$-adic integers, your sum would be $4+5=0+2\cdot 3 = 6$. You need to get $9$, the same as with normal addition. –  Thomas Andrews Jul 25 '12 at 15:40
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Defining $p$-adic multiplication using its base $p$ representation is at least as hard as defining multiplication in $\mathbb Z$ by the algorithm of long multiplication. It is essentially the same logic, only with infinitely many "digits." –  Thomas Andrews Jul 25 '12 at 16:44
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2 Answers

up vote 6 down vote accepted

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: why do we ever work with the integers instead of the rationals? Because sometimes it's the integers that are relevant to a problem and not the rationals. (By the way, there are plenty of sums with "rational" limits. For example, $\frac{1}{4} = \frac{1}{1 + 3}$ exists in $\mathbb{Z}_3$.) –  Qiaochu Yuan Jul 25 '12 at 15:14
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@Clark: in any case, the topological properties of $\mathbb{Z}_p$ and $\mathbb{Q}_p$ do matter ($p$-adic convegence is important as you can see from the above). Both are topological rings (a mixture of topology and algebra) and the way in which the two structures interact is important. –  Qiaochu Yuan Jul 25 '12 at 15:15
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@Clark: $\frac{1}{1 + 3} = 1 - 3 + 9 - 27 \pm ...$. The addition in $\mathbb{Z}_3$ is not componentwise addition $\bmod 3$; you need to carry just like you would in base $3$. –  Qiaochu Yuan Jul 25 '12 at 15:25
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Also, even algebraically, there is a notion of "limits." There is a sense in which $\mathbb Z_p$ is the "limit" of the rings $\mathbb Z/p^k\mathbb Z$. (Actually, it is the "inverse limit" of these rings.) There is no such inverse limit definition for $\mathbb Q_p$, however. –  Thomas Andrews Jul 25 '12 at 15:25
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@Clark: the Euclidean norm cannot be extended to $\mathbb{Z}_p$. –  Qiaochu Yuan Jul 25 '12 at 16:40
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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. –  Thomas Andrews Jul 25 '12 at 17:20
    
Thanks a lot Dylan!! –  Matt N. Jul 25 '12 at 18:43
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