1Basic theory

III Local Fields

1.3 Topological rings

Recall that we previously constructed the valuation ring

O

K

. Since the valued

field

K

itself has a topology, the valuation ring inherits a subspace topology.

This is in fact a ring topology.

Definition

(Topological ring)

.

Let

R

be a ring. A topology on

R

is called a

ring topology if addition and multiplication are continuous maps

R × R → R

. A

ring with a ring topology is a topological ring.

Example. R

and

C

with the usual topologies and usual ring structures are

topological rings.

Exercise.

Let

K

be a valued field. Then

K

is a topological ring. We can see

this from the fact that the product topology on

K ×K

is induced by the metric

d((x

0

, y

0

), (x

1

, y

1

)) = max(|x

0

− x

1

|, |y

0

− y

1

|).

Now if we are just randomly given a ring, there is a general way of constructing

a ring topology. The idea is that we pick an ideal

I

and declare its elements to

be small. For example, in a valued ring, we can pick

I

=

{x ∈ O

K

:

|x| <

1

}

.

Now if you are not only in

I

, but

I

2

, then you are even smaller. So we have a

hierarchy of small sets

I ⊇ I

2

⊇ I

3

⊇ I

4

⊇ ···

Now to make this a topology on

R

, we say that a subset

U ⊆ R

is open if every

x ∈ U

is contained in some translation of

I

n

(for some

n

). In other words, we

need some y ∈ R such that

x ∈ y + I

n

⊆ U.

But since

I

n

is additively closed, this is equivalent to saying

x

+

I

n

⊆ U

. So we

make the following definition:

Definition

(

I

-adically open)

.

Let

R

be a ring and

I ⊆ R

an ideal. A subset

U ⊆ R

is called

I

-adically open if for all

x ∈ U

, there is some

n ≥

1 such that

x + I

n

⊆ U.

Proposition.

The set of all

I

-adically open sets form a topology on

R

, called

the I-adic topology.

Note that the

I

-adic topology isn’t really the kind of topology we are used

to thinking about, just like the topology on a valued field is also very weird.

Instead, it is a “filter” for telling us how small things are.

Proof.

By definition, we have

∅

and

R

are open, and arbitrary unions are clearly

open. If

U, V

are

I

-adically open, and

x ∈ U ∩ V

, then there are

n, m

such that

x + I

n

⊆ U and x + I

m

⊆ V . Then x + I

max(m,n)

⊆ U ∩ V .

Exercise. Check that the I-adic topology is a ring topology.

In the special case where

I

=

xR

, we often call the

I

-adic topology the

x

-adic

topology.

Now we want to tackle the notion of completeness. We will consider the case

of I = xR for motivation, but the actual definition will be completely general.

If we pick the

x

-adic topology, then we are essentially declaring that we take

x to be small. So intuitively, we would expect power series like

a

0

+ a

1

x + a

2

x

2

+ a

3

x

3

+ ···

to “converge”, at least if the

a

i

are “of bounded size”. In general, the

a

i

are

“not too big” if

a

i

x

i

is genuinely a member of

x

i

R

, as opposed to some silly thing

like x

−i

.

As in the case of analysis, we would like to think of these infinite series as a

sequence of partial sums

(a

0

, a

0

+ a

1

x, a

0

+ a

1

x + a

2

x

2

, ···)

Now if we denote the limit as

L

, then we can think of this sequence alternatively

as

(L mod I, L mod I

2

, L mod I

3

, ···).

The key property of this sequence is that if we take

L mod I

k

and reduce it mod

I

k−1

, then we obtain L mod I

k−1

.

In general, suppose we have a sequence

(b

n

∈ R/I

n

)

∞

n=1

.

such that

b

n

mod I

n−1

=

b

n−1

. Then we want to say that the ring is

I

-adically

complete if every sequence satisfying this property is actually of the form

(L mod I, L mod I

2

, L mod I

3

, ···)

for some

L

. Alternatively, we can take the

I

-adic completion to be the collection

of all such sequences, and then a space is

I

-adically complete it is isomorphic to

its I-adic completion.

To do this, we need to build up some technical machinery. The kind of

sequences we’ve just mentioned is a special case of an inverse limit.

Definition

(Inverse/projective limit)

.

Let

R

1

, R

2

, , ···

be topological rings, with

continuous homomorphisms f

n

: R

n+1

→ R

n

.

R

1

R

2

R

3

R

4

···

f

1

f

2

f

3

The inverse limit or projective limit of the R

i

is the ring

lim

←−

R

n

=

(

(x

n

) ∈

Y

n

R

n

: f

n

(x

n+1

) = x

n

)

,

with coordinate-wise addition and multiplication, together with the subspace

topology coming from the product topology of

Q

R

n

. This topology is known as

the inverse limit topology.

Proposition. The inverse limit topology is a ring topology.

Proof sketch. We can fit the addition and multiplication maps into diagrams

lim

←−

R

n

× lim

←−

R

n

lim

←−

R

n

Q

R

n

×

Q

R

n

Q

R

n

By the definition of the subspace topology, it suffices to show that the cor-

responding maps on

Q

R

n

are continuous. By the universal property of the

product, it suffices to show that the projects

Q

R

n

×

Q

R

n

→ R

m

is continuous

for all

m

. But this map can alternatively be obtained by first projecting to

R

m

,

then doing multiplication in

R

m

, and projection is continuous. So the result

follows.

It is easy to see the following universal property of the inverse limit topology:

Proposition.

Giving a continuous ring homomorphism

g

:

S → lim

←−

R

n

is the

same as giving a continuous ring homomorphism

g

n

:

S → R

n

for each

n

, such

that each of the following diagram commutes:

S R

n

R

n−1

g

n

g

n−1

f

n−1

Definition

(

I

-adic completion)

.

Let

R

be a ring and

I

be an ideal. The

I

-adic

completion of R is the topological ring

lim

←−

R/I

n

,

where

R/I

n

has the discrete topology, and

R/I

n+1

→ R/I

n

is the quotient map.

There is an evident map

ν : R → lim

←−

R/I

n

r 7→ (r mod I

n

)

.

This map is a continuous ring homomorphism if

R

is given the

I

-adic topology.

Definition

(

I

-adically complete)

.

We say that

R

is

I

-adically complete if

ν

is a

bijection.

Exercise. If ν is a bijection, then ν is in fact a homeomorphism.