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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.2012-07-25
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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.2012-07-25
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    @ThomasAndrews Thank you Thomas!2012-07-25
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    I guess what I don't like about this definition is that it makes it hard to see how the addition and multiplication work.2012-07-25
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    @DylanMoreland Is it not just the same as it is for power series?2012-07-25
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    @ClarkKent This was my fear. Could you be more specific as to how you would, say, add two of these things? It is certainly not true that $\mathbb Z_p \simeq \mathbb F_p[[X]]$. The "coefficients" interact, even under addition.2012-07-25
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    @DylanMoreland For addition: $$ \sum_{k=0}^\infty a_k p^k + \sum_{k=0}^\infty b_k p^k = \sum_{k=0}^\infty (a_k + b_k) p^k$$ where we take $a_k + b_k$ to be $\mod p$.2012-07-25
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    @DylanMoreland For multiplication: $$ \sum_{k=0}^\infty a_k p^k \cdot \sum_{i=0}^\infty b_i p^i = \sum_{k=0}^\infty \sum_{i=0}^k (a_{k-i} b_i) p^k$$ where again we take the new coefficients $\mod p$.2012-07-25
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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.)2012-07-25
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    @ThomasAndrews Wait but if we add two sums each of which is zero almost everywhere then surely the sum is also zero almost everywhere.2012-07-25
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    And that definition of multiplication doesn't work, either, since either $\sum_{i=0}^k a_{k-i}b_i$ is not always in $\{0,...,p-1\}$ or you are computing $\pmod p$, which would be wrong.2012-07-25
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    @ThomasAndrews Yes, I meant $\mod p$ again, like in the case of addition. I edited the comment.2012-07-25
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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.2012-07-25
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    @ThomasAndrews Thank you. I didn't realise. I'm trying to figure out how to fix addition and multiplication now as I can't seem to find the correct definitions on the [Wikipedia page](http://en.wikipedia.org/wiki/P-adic_number).2012-07-25
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    @ClarkKent It is actually much easier to define multiplication using the "inverse limit" definition of $\mathbb Z_p$2012-07-25
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    @ThomasAndrews Ok, fixing multiplication into a definition that works might be not so easy but it seems to me that addition should be feasible without inverse limits.2012-07-25
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    It is certainly possible to do addition without it. Essentially, you are doing the usual "long addition" in base $p$. Add $a_0+b_0$. If the result is more than $p$ then $c_0=a_0+b_0-p$ and you "carry the one." Otherwise, your get $c_0=a_0+b_0$ and you "carry zero." To get $c_1$, you start with $a_1+b_1+X$ where $X$ is the "carry" from the zero point of view, and again you determine if $a_1+b_1+X$ is bigger than $X$ and define $c_1$ depending on that comparison, and get another carry number (0 or 1) which you need to use to compute $c_2$, etc. But even here, the inverse limit is easier2012-07-25
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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."2012-07-25
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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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    For some reason my eyes skipped over "Those $p$-adic numbers for which $a_i = 0$ for all $i < 0$ are also called the $p$-adic integers." [here](http://en.wikipedia.org/wiki/P-adic_number#p-adic_expansions). Why we have to exclude negative power is not clear to me though. Perhaps because otherwise we could have a sum with a rational limit? But that shouldn't be a problem because we're in algebra world where topological properties of spaces don't matter...2012-07-25
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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 $\mathbb{Z}_3$.)2012-07-25
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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.2012-07-25
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    @Clark: this "have to" terminology is mysterious to me and I think it comes from spending too much time dealing with math teachers who say that things have to be a certain way and not another way. Things do not have to be _any way_ in mathematics. We _choose_ things to be certain ways to provide ourselves with a convenient language to describe and explore interesting phenomena. These are choices that we collectively make, not commandments delivered from on high.2012-07-25
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    It turns out that given any sequence of $a_k\in\mathbb Z_p$, the sum $\sum_{k=m}^\infty a_kp^k$ converges in $\mathbb Q_p$, but that is not true of we take $a_k\in\mathbb Q_p$. For example, $a_k=p^{-k}$ will not yield a convergent series. That's just one of the reasons we distinguish $\mathbb Z_p$ and $\mathbb Q_p$.2012-07-25
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    Regarding your first comment: If $$\mathbb Z_3 = \{ \sum_{k=0}^\infty a_k 3^k \mid a_k \in \{0, 1, 2\} \} $$ how can we have $\frac{1}{1 + 3}$ in $\mathbb Z_3$? Did you mean to write $\mathbb Q_3$ there?2012-07-25
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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 $3$.2012-07-25
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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.2012-07-25
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    @ThomasAndrews The reason why I am reading about $p$-adics is because my plan is to next move on to inverse limits : )2012-07-25
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    QiaochuYuan @your last comment: Oh my, of course. I did not see that but it's clear now. Thank you!2012-07-25
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    @Qiaochu Well, $\frac{1}{1 - pz} $ converges even if $|pz|$ is not strictly less than $1$ because you (sneakily) swapped the Euclidean norm $|\cdot|$ for the $p$-adic norm $|\cdot|_p$, right? Is there a name for the thing we get if we keep the Euclidean norm?2012-07-25
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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