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.