1Basic theory

III Local Fields

1.2 Rings

Definition

(Integral element)

.

Let

R ⊆ S

be rings and

s ∈ S

. We say

s

is

integral over R if there is some monic f ∈ R[x] such that f(s) = 0.

Example. Any r ∈ R is integral (take f(x) = x − r).

Example.

Take

Z ⊆ C

. Then

z ∈ C

is integral over

Z

if it is an algebraic

integer (by definition of algebraic integer). For example,

√

2

is an algebraic

integer, but

1

√

2

is not.

We would like to prove the following characterization of integral elements:

Theorem.

Let

R ⊆ S

be rings. Then

s

1

, ··· , s

n

∈ S

are all integral iff

R[s

1

, ··· , s

n

] ⊆ S is a finitely-generated R-module.

Note that

R

[

s

1

, ··· , s

n

] is by definition a finitely-generated

R

-algebra, but

requiring it to be finitely-generated as a module is stronger.

Here one direction is easy. It is not hard to show that if

s

1

, ··· , s

n

are all

integral, then

R

[

s

1

, ··· , s

n

] is finitely-generated. However to show the other

direction, we need to find some clever trick to produce a monic polynomial that

kills the s

i

.

The trick we need is the adjugate matrix we know and love from IA Vectors

and Matrices.

Definition

(Adjoint/Adjugate matrix)

.

Let

A

= (

a

ij

) be an

n × n

matrix with

coefficients in a ring

R

. The adjugate matrix or adjoint matrix

A

∗

= (

a

∗

ij

) of

A

is defined by

a

∗

ij

= (−1)

i+j

det(A

ij

),

where

A

ij

is an (

n −

1)

×

(

n −

1) matrix obtained from

A

by deleting the

i

th

column and the jth row.

As we know from IA, the following property holds for the adjugate matrix:

Proposition.

For any

A

, we have

A

∗

A

=

AA

∗

=

det

(

A

)

I

, where

I

is the

identity matrix.

With this, we can prove our claim:

Proof of theorem. Note that we can construct R[s

1

, ··· , s

n

] by a sequence

R ⊆ R[s

1

] ⊆ R[s

1

, s

2

] ⊆ ··· ⊆ R[s

1

, ··· , s

n

] ⊆ S,

and each

s

i

is integral over

R

[

s

1

, ··· , s

n−1

]. Since the finite extension of a finite

extension is still finite, it suffices to prove it for the case

n

= 1, and we write

s

for s

1

.

Suppose

f

(

x

)

∈ R

[

x

] is monic such that

f

(

s

) = 0. If

g

(

x

)

∈ R

[

x

], then there

is some

q, r ∈ R

[

x

] such that

g

(

x

) =

f

(

x

)

q

(

x

) +

r

(

x

) with

deg r < deg f

. Then

g

(

s

) =

r

(

s

). So any polynomial expression in

s

can be written as a polynomial

expression with degree less than

deg f

. So

R

[

s

] is generated by 1

, s, ··· , s

deg f −1

.

In the other direction, let

t

1

, ··· , t

d

be

R

-module generators of

R

[

s

1

, ··· , s

n

].

We show that in fact any element of

R

[

s

1

, ··· , s

n

] is integral over

R

. Consider

any element b ∈ R[s

1

, ··· , s

n

]. Then there is some a

ij

∈ R such that

bt

i

=

d

X

j=1

a

ij

t

j

.

In matrix form, this says

(bI − A)t = 0.

We now multiply by (bI − A)

∗

to obtain

det(bI − A)t

j

= 0

for all j. Now we know 1 ∈ R. So 1 =

P

c

j

t

j

for some c

j

∈ R. Then we have

det(bI − A) = det(bI − A)

X

c

j

t

j

=

X

c

j

(det(bI − A)t

j

) = 0.

Since det(bI − A) is a monic polynomial in b, it follows that b is integral.

Using this characterization, the following result is obvious:

Corollary.

Let

R ⊆ S

be rings. If

s

1

, s

2

∈ S

are integral over

R

, then

s

1

+

s

2

and

s

1

s

2

are integral over

R

. In particular, the set

˜

R ⊆ S

of all elements in

S

integral over R is a ring, known as the integral closure of R in S.

Proof.

If

s

1

, s

2

are integral, then

R

[

s

1

, s

2

] is a finite extension over

R

. Since

s

1

+ s

2

and s

1

s

2

are elements of R[s

1

, s

2

], they are also integral over R.

Definition

(Integrally closed)

.

Given a ring extension

R ⊆ S

, we say

R

is

integrally closed in S if

˜

R = R.