I think this post is more than two years late in coming, but anywhow…
This post introduces the
-adic integers
, and the
-adic numbers
. The one-sentence description is that these are “integers/rationals carrying full mod
information” (and only that information).
The first four sections will cover the founding definitions culminating in a short solution to a USA TST problem.
In this whole post,
is always a prime. Much of this is based off of Chapter 3A from Straight from the Book.
1. Motivation
Before really telling you what
and
are, let me tell you what you might expect them to do.
In elementary/olympiad number theory, we’re already well-familiar with the following two ideas:
- Taking modulo a prime
or prime
, and
- Looking at the exponent
.
Let me expand on the first point. Suppose we have some Diophantine equation. In olympiad contexts, one can take an equation modulo
to gain something else to work with. Unfortunately, taking modulo
loses some information: (the reduction
is far from injective).
If we want finer control, we could consider instead taking modulo
, rather than taking modulo
. This can also give some new information (cubes modulo
, anyone?), but it has the disadvantage that
isn’t a field, so we lose a lot of the nice algebraic properties that we got if we take modulo
.
One of the goals of
-adic numbers is that we can get around these two issues I described. The
-adic numbers we introduce is going to have the following properties:
- You can “take modulo
for all
at once”. In olympiad contexts, we are used to picking a particular modulus and then seeing what happens if we take that modulus. But with
-adic numbers, we won’t have to make that choice. An equation of
-adic numbers carries enough information to take modulo
.
- The numbers
form a field, the nicest possible algebraic structure:
makes sense. Contrast this with
, which is not even an integral domain.
- It doesn’t lose as much information as taking modulo
does: rather than the surjective
we have an injective map
.
- Despite this, you “ignore” some “irrelevant” data. Just like taking modulo
, you want to zoom-in on a particular type of algebraic information, and this means necessarily losing sight of other things. (To draw an analogy: the equation
has no integer solutions, because, well, squares are nonnegative. But you will find that this equation has solutions modulo any prime
, because once you take modulo
you stop being able to talk about numbers being nonnegative. The same thing will happen if we work in
-adics: the above equation has a solution in
for every prime
.)
So, you can think of
-adic numbers as the right tool to use if you only really care about modulo
information, but normal
isn’t quite powerful enough.
To be more concrete, I’ll give a poster example now:
Example 1 (USA TST 2002/2)
For a prime
, show the value of

does not depend on
.
Here is a problem where we clearly only care about
-type information. Yet it’s a nontrivial challenge to do the necessary manipulations mod
(try it!). The basic issue is that there is no good way to deal with the denominators modulo
(in part
is not even an integral domain).
However, with
-adic analysis we’re going to be able to overcome these limitations and give a “straightforward” proof by using the identity

Such an identity makes no sense over
or
for converge reasons, but it will work fine over the
, which is all we need.
2. Algebraic perspective
We now construct
and
. I promised earlier that a
-adic integer will let you look at “all residues modulo
” at once. This definition will formalize this.
2.1. Definition of 
Example 3 (Some
-adic integers)
Let
. Every usual integer
generates a (compatible) sequence of residues modulo
for each
, so we can view each ordinary integer as
-adic one:

On the other hand, there are sequences of residues which do not correspond to any usual integer despite satisfying compatibility relations, such as

which can be thought of as
.
In this way we get an injective map

which is not surjective. So there are more
-adic integers than usual integers.
(Remark for experts: those of you familiar with category theory might recognize that this definition can be written concisely as

where the inverse limit is taken across
.)
Exercise 4
Check that
is an integral domain.
2.2. Base
expansion
Here is another way to think about
-adic integers using “base
”. As in the example earlier, every usual integer can be written in base
, for example

More generally, given any
, we can write down a “base
” expansion in the sense that there are exactly
choices of
given
. Continuing the example earlier, we would write

and in general we can write

where
, such that the equation holds modulo
for each
. Note the expansion is infinite to the left, which is different from what you’re used to.
(Amusingly, negative integers also have infinite base
expansions:
, corresponding to
.)
Thus you may often hear the advertisement that a
-adic integer is an “possibly infinite base
expansion”. This is correct, but later on we’ll be thinking of
in a more and more “analytic” way, and so I prefer to think of this as a “Taylor series with base
”. Indeed, much of your intuition from generating functions
(where
is a field) will carry over to
.
2.3. Constructing 
Here is one way in which your intuition from generating functions carries over:
Proposition 5 (Non-multiples of
are all invertible)
The number
is invertible if and only if
. In symbols,

Contrast this with the corresponding statement for
: a generating function
is invertible iff
.
Proof: If
then
, so clearly not invertible. Otherwise,
for all
, so we can take an inverse
modulo
, with
. As the
are themselves compatible, the element
is an inverse. 
Example 6 (We have
)
We claim the earlier example is actually

Indeed, multiplying it by
gives

(Compare this with the “geometric series”
. We’ll actually be able to formalize this later, but not yet.)
With this observation, here is now the definition of
.
Definition 8 (Introducing
)
Since
is an integral domain, we let
denote its field of fractions. These are the
-adic numbers.
Continuing our generating functions analogy:
![\displaystyle \mathbb Z_p \text{ is to } \mathbb Q_p \quad\text{as}\quad K[[X]] \text{ is to } K((X)). \displaystyle \mathbb Z_p \text{ is to } \mathbb Q_p \quad\text{as}\quad K[[X]] \text{ is to } K((X)).](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmathbb+Z_p+%5Ctext%7B+is+to+%7D+%5Cmathbb+Q_p+%5Cquad%5Ctext%7Bas%7D%5Cquad+K%5B%5BX%5D%5D+%5Ctext%7B+is+to+%7D+K%28%28X%29%29.+&bg=ffffff&fg=000000&s=0)
This means
is “Laurent series with base
”, and in particular according to the earlier proposition we deduce:
Proposition 9 (
looks like formal Laurent series)
Every nonzero element of
is uniquely of the form

Thus, continuing our base
analogy, elements of
are in bijection with “Laurent series”

for
. So the base
representations of elements of
can be thought of as the same as usual, but extending infinitely far to the left (rather than to the right).
(Fair warning: the field
has characteristic zero, not
.)
(At this point I want to make a remark about the fact
, connecting it to the wish-list of properties I had before. In elementary number theory you can take equations modulo
, but if you do the quantity
doesn’t make sense unless you know
. You can’t fix this by just taking modulo
since then you need
to get
, ad infinitum. You can work around issues like this, but the nice feature of
and
is that you have modulo
information for “all
at once”: the information of
packages all the modulo
information simultaneously. So you can divide by
with no repercussions.)
3. Analytic perspective
3.1. Definition
Up until now we’ve been thinking about things mostly algebraically, but moving forward it will be helpful to start using the language of analysis. Usually, two real numbers are considered “close” if they are close on the number of line, but for
-adic purposes we only care about modulo
information. So, we’ll instead think of two elements of
or
as “close” if they differ by a large multiple of
.
For this we’ll borrow the familiar
from elementary number theory.
This fulfills the promise that
and
are close if they look the same modulo
for large
; in that case
is large and accordingly
is small.
3.2. Ultrametric space
In this way,
and
becomes a metric space with metric given by
.
Exercise 12
Suppose
is continuous and
for every
. Prove that
.
In fact, these spaces satisfy a stronger form of the triangle inequality than you are used to from
.
Proposition 13 (
is an ultrametric)
For any
, we have the strong triangle inequality

Equality holds if (but not only if)
.
However,
is more than just a metric space: it is a field, with its own addition and multiplication. This means we can do analysis just like in
or
: basically, any notion such as “continuous function”, “convergent series”, et cetera has a
-adic analog. In particular, we can define what it means for an infinite sum to converge:
With this definition in place, the “base
” discussion we had earlier is now true in the analytic sense: if
then

Indeed, the
th partial sum is divisible by
, hence the partial sums approach
as
.
While the definitions are all the same, there are some changes in properties that should be true. For example, in
convergence of partial sums is simpler:
Proposition 15 (
iff convergence of series)
A series
in
converges to some limit if and only if
.
Contrast this with
in
. You can think of this as a consequence of strong triangle inequality. Proof: By multiplying by a large enough power of
, we may assume
. (This isn’t actually necessary, but makes the notation nicer.)
Observe that
must eventually stabilize, since for large enough
we have
. So let
be the eventual residue modulo
of
for large
. In the same way let
be the eventual residue modulo
, and so on. Then one can check we approach the limit
. 
Here’s a couple exercises to get you used to thinking of
and
as metric spaces.
Exercise 16 (
is compact)
Show that
is not compact, but
is. (For the latter, I recommend using sequential continuity.)
Exercise 17 (Totally disconnected)
Show that both
and
are totally disconnected: there are no connected sets other than the empty set and singleton sets.
3.3. More fun with geometric series
While we’re at it, let’s finally state the
-adic analog of the geometric series formula.
Proposition 18 (Geometric series)
Let
with
. Then

Proof: Note that the partial sums satisfy
, and
as
since
. 
So,
is really a correct convergence in
. And so on.
If you buy the analogy that
is generating functions with base
, then all the olympiad generating functions you might be used to have
-adic analogs. For example, you can prove more generally that:
Theorem 19 (Generalized binomial theorem)
If
and
, then for any
we have the series convergence

(I haven’t defined
, but it has the properties you expect.) The proof is as in the real case; even the theorem statement is the same except for the change for the extra subscript of
. I won’t elaborate too much on this now, since
-adic exponentiation will be described in much more detail in the next post.
3.4. Completeness
Note that the definition of
could have been given for
as well; we didn’t need
to introduce it (after all, we have
in olympiads already). The big important theorem I must state now is:
Theorem 20 (
is complete)
The space
is the completion of
with respect to
.
This is the definition of
you’ll see more frequently; one then defines
in terms of
(rather than vice-versa) according to

(Remark for experts:
is a field with
a non-Arcihmedian valuation; then
is its valuation ring.)
Let me justify why this definition is philosophically nice.
Suppose you are a numerical analyst and you want to estimate the value of the sum

to within
. The sum
consists entirely of rational numbers, so the problem statement would be fair game for ancient Greece. But it turns out that in order to get a good estimate, it really helps if you know about the real numbers: because then you can construct the infinite series
, and deduce that
, up to some small error term from the terms past
, which can be bounded.
Of course, in order to have access to enough theory to prove that
, you need to have the real numbers; it’s impossible to do serious analysis in the non-complete space
, where e.g. the sequence
,
,
,
, \dots is considered “not convergent” because
. Instead, all analysis is done in the completion of
, namely
.
Now suppose you are an olympiad contestant and want to estimate the sum

to within mod
(i.e. to within
in
). Even though
is a rational number, it still helps to be able to do analysis with infinite sums, and then bound the error term (i.e. take mod
). But the space
is not complete with respect to
either, and thus it makes sense to work in the completion of
with respect to
. This is exactly
.
4. Solving USA TST 2002/2
Let’s finally solve Example~1, which asks to compute

Armed with the generalized binomial theorem, this becomes straightforward.

Using the elementary facts that
and
, this solves the problem.