1Basic theory
III Local Fields
1.4 The padic 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
adic numbers,
and study some of its basic properties.
Let
x ∈ Q
be nonzero. 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
adic absolute value)
.
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 padic 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
−m−n
= x
p
y
p
.
So this is multiplicative. Finally, we have
x + y
p
=
p
n
ab + p
m−n
cb
bd
≤ p
−n
= max(x
p
, y
p
).
Note that we must have
bd
coprime to
p
, but
ab
+
p
m−n
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
adic numbers)
.
The
p
adic numbers
Q
p
is the completion of
Q
with respect to  · 
p
.
Definition (padic integers). The valuation ring
Z
p
= {x ∈ Q
p
: x
p
≤ 1}
is the padic integers.
Proposition. Z
p
is the closure of Z inside Q
p
.
Proof. If x ∈ Z is nonzero, 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 nonzero 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
adic topology (i.e.
the pZadic topology).
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 padically complete and is (isomorphic to) the padic 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
=
n−1
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
+ ··· .