1Basic theory

III Local Fields

1.4 The p-adic numbers
For the rest of this course, p is going to be a prime number.
We consider a particular case of valued fields, namely the
p
and study some of its basic properties.
Let
x Q
be non-zero. Then by uniqueness of factorization, we can write
x
uniquely as
x = p
n
a
b
,
where a, b, n Z, b > 0 and a, b, p are pairwise coprime.
Definition
(
p
.
The
p
-adic absolute value on
Q
is the
function | · |
p
: Q R
0
given by
|x|
p
=
(
0 x = 0
p
n
x = p
n
a
b
as above
.
Proposition. The p-adic absolute value is an absolute value.
Proof. It is clear that |x|
p
= 0 iff x = 0.
Suppose we have
x = p
n
a
b
, y = p
m
c
d
.
We wlog m n. Then we have
|xy|
p
=
p
n+m
ac
bd
= p
mn
= |x|
p
|y|
p
.
So this is multiplicative. Finally, we have
|x + y|
p
=
p
n
ab + p
mn
cb
bd
p
n
= max(|x|
p
, |y|
p
).
Note that we must have
bd
coprime to
p
, but
ab
+
p
mn
cb
need not be. However,
any extra powers of
p
could only decrease the absolute value, hence the above
result.
Note that if x Z is an integer, then |x|
p
= p
n
iff p
n
|| x (we say p
n
|| x if
p
n
| x and p
n+1
- x).
Definition
(
p
.
The
p
Q
p
is the completion of
Q
with respect to | · |
p
.
Definition (p-adic integers). The valuation ring
Z
p
= {x Q
p
: |x|
p
1}
is the p-adic integers.
Proposition. Z
p
is the closure of Z inside Q
p
.
Proof. If x Z is non-zero, then x = p
n
a with n 0. So |x|
p
1. So Z Z
p
.
We now want to show that Z is dense in Z
p
. We know the set
Z
(p)
= {x Q : |x|
p
1}
is dense inside
Z
p
, essentially by definition. So it suffices to show that
Z
is dense
in Z
(p)
. We let x Z
(p)
\ {0}, say
x = p
n
a
b
, n 0.
It suffices to find x
i
Z such that x
i
1
b
. Then we have p
n
ax
i
x.
Since (
b, p
) = 1, we can find
x
i
, y
i
Z
such that
bx
i
+
p
i
y
i
= 1 for all
i
1.
So
x
i
1
b
p
=
1
b
p
|bx
i
1|
p
= |p
i
y
i
|
p
p
i
0.
So done.
Proposition. The non-zero ideals of Z
p
are p
n
Z
p
for n 0. Moreover,
Z
p
n
Z
=
Z
p
p
n
Z
p
.
Proof.
Let 0
6
=
I Z
p
be an ideal, and pick
x I
such that
|x|
p
is maximal.
This supremum exists and is attained because the possible values of the absolute
values are discrete and bounded above. If
y I
, then by maximality, we have
|y|
p
|x|
p
. So we have
|yx
1
|
p
1. So
yx
1
Z
p
, and this implies that
y
= (
yx
1
)
x xZ
p
. So
I xZ
p
, and we obviously have
xZ
p
I
. So we have
I = xZ
p
.
Now if
x
=
p
n
a
b
, then since
a
b
is invertible in
Z
p
, we have
xZ
p
=
p
n
Z
p
. So
I = p
n
Z
p
.
To show the second part, consider the map
f
n
: Z
Z
p
p
n
Z
p
given by the inclusion map followed by quotienting. Now
p
n
Z
p
=
{x
:
|x|
p
p
n
.
So we have
ker f
n
= {x Z : |x|
p
p
n
} = p
n
Z.
Now since
Z
is dense in
Z
p
, we know the image of
f
n
is dense in
Z
p
/p
n
Z
p
.
But
Z
p
/p
n
Z
p
has the discrete topology. So
f
n
is surjective. So
f
n
induces an
isomorphism Z/p
n
Z
=
Z
p
/p
n
Z
p
.
Corollary. Z
p
is a PID with a unique prime element p (up to units).
This is pretty much the point of the
p
-adic numbers there are a lot of
primes in Z, and by passing on to Z
p
, we are left with just one of them.
Proposition.
The topology on
Z
induced by
| · |
p
is the
p
Proof.
Let
U Z
. By definition,
U
is open wrt
| · |
p
iff for all
x U
, there is
an n N such that
{y Z : |y x|
p
p
n
} U.
On the other hand,
U
is open in the
p
-adic topology iff for all
x U
, there is
some n 0 such that x + p
n
Z U. But we have
{y Z : |y x|
p
p
n
} = x + p
n
Z.
So done.
Proposition. Z
p
is p-adically complete and is (isomorphic to) the p-adic com-
pletion of Z.
Proof. The second part follows from the first as follows: we have the maps
Z
p
lim
Z
p
/(p
n
Z
p
) lim Z/(p
n
Z)
ν
(f
n
)
n
We know the map induced by (
f
n
)
n
is an isomorphism. So we just have to show
that ν is an isomorphism
To prove the first part, we have
x ker ν
iff
x p
n
Z
p
for all
n
iff
|x|
p
p
n
for all n iff |x|
p
= 0 iff x = 0. So the map is injective.
To show surjectivity, we let
z
n
lim
Z
p
/p
n
Z
p
.
We define a
i
{0, 1, ··· , p 1} recursively such that
x
n
=
n1
X
i=0
a
i
p
i
is the unique representative of
z
n
in the set of integers
{
0
,
1
, ··· , p
n
1
}
. Then
x =
X
i=0
a
i
p
i
exists in
Z
p
and maps to
x x
n
z
n
(
mod p
n
) for all
n
0. So
ν
(
x
) = (
z
n
).
So the map is surjective. So ν is bijective.
Corollary. Every a Z
p
has a unique expansion
a =
X
i=0
a
i
p
i
.
with a
i
{0, ··· , p 1}.
More generally, for any a Q
×
, there is a unique expansion
a =
X
i=n
a
i
p
i
for a
i
{0, ··· , p 1}, a
n
6= 0 and
n = log
p
|a|
p
Z.
Proof.
The second part follows from the first part by multiplying
a
by
p
n
.
Example. We have
1
1 p
= 1 + p + p
2
+ p
3
+ ··· .