# 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.

• 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. Commented Jul 25, 2012 at 15:06
• "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. Commented Jul 25, 2012 at 15:10
• @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.) Commented Jul 25, 2012 at 15:37
• @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. Commented Jul 25, 2012 at 15:40
• 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." Commented Jul 25, 2012 at 16:44

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.

• 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. Commented Jul 25, 2012 at 17:20

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.)

• @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 $\mathbb{Z}_3$.) Commented Jul 25, 2012 at 15:14
• @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. Commented Jul 25, 2012 at 15:15
• @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. Commented Jul 25, 2012 at 15:18
• @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 $3$. Commented Jul 25, 2012 at 15:25
• 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. Commented Jul 25, 2012 at 15:25