PMATH 441/641 ALGEBRAIC NUMBER THEORY
1.
Lecture: Wednesday, January 5, 2000
NO TEXT
References:
•
Number Fields (Marcus)
•
Algebraic Number Theory (Lang) (Stewart & Hill)
Marks:
•
Final Exam 65%
•
Midterm 25%
•
Assignments 10%
Definition.
An
algebraic integer
is the root of a monic polynomial in
Z
[
x
].
An
algebraic number
is the root of any nonzero polynomial in
Z
[
x
].
We are interested in studying the structure of the ring of algebraic integers in an algebraic
number field. A number field is a finite extension of
Q
. We’ll assume that the number fields
we consider are all subfields of
C
.
Definition.
Suppose that
K
and
L
are fields with
K
⊆
L
. Then
K
is a
subfield
of
L
and
L
is an
extension field
of
K
.
We denote the dimension of
L
as a vector space over
K
by [
L
:
K
] .
If [
L
:
K
]
<
∞
, we say
L
is a
finite extension
of
K
.
Definition.
Suppose that
H
is a field with
K
⊆
H
⊆
L
. Then we say
H
is an
intermediate
field
of
K
and
L
. Recall that [
L
:
K
] = [
L
:
H
][
H
:
K
].
Definition.
A polynomial
f
∈
K
[
x
] is said to be
irreducible
over
K
iff whenever
f
=
gh
with
g, h
∈
K
[
x
], we have
g
or
h
constant.
Recall that
K
[
x
] is a Principal Ideal Domain.
Definition.
Let
K
be a subfield of
C
and let
θ
∈
C
be an algebraic number. We denote
by
K
(
θ
) the smallest subfield of
C
containing
K
and
θ
,
Definition.
Let
K
be a subfield of
C
and let
θ
∈
C
to be algebraic over
K
. A polynomial
in
K
[
x
] is said to be a
minimal polynomial
of
θ
over
K
if it is monic, has
θ
as a root, and
has degree as small as possible with these properties.
Theorem 1.
Let
K
⊆
C
. Let
θ
∈
C
be algebraic over
K
. Then there is a unique minimal
polynomial of
θ
over
K
.
Proof:
Plainly there is at least one. Suppose that
p
1
(
x
) and
p
2
(
x
) are minimal polynomials for
θ
over
K
.
Consider
p
1
(
x
)

p
2
(
x
). Since
p
1
, p
2
are monic and of minimal degree, the degree of
p
1
(
x
)

1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
PMATH 441/641 ALGEBRAIC NUMBER THEORY
p
2
(
x
) is strictly smaller than the degree of
p
1
(
x
), or
p
1
(
x
) =
p
2
(
x
).
In the former case, we contradict the minimality of the degree since
p
1
(
θ
)

p
2
(
θ
) = 0. Thus
p
1
=
p
2
and the result follows.
Thus we can speak of “the” minimal polynomial of
θ
over
K
.
Definition.
Let
K
⊆
C
. Let
θ
be algebraic over
K
. The
degree of
θ
over
K
is the degree
of the minimal polynomial of
θ
over
K
.
Remark.
Let
K
∈
C
and
θ
∈
C
with
θ
algebraic over
K
.
Let
p
(
x
) be the minimal
polynomial of
θ
over
K
. Suppose
f
∈
K
[
x
] with
f
(
θ
) = 0. Then
p

f
in
K
[
x
]. To see this
note that by the Division Algorithm in
K
[
x
],
f
(
x
) =
q
(
x
)
p
(
x
) +
r
(
x
) with
r
= 0 or deg
r <
deg
p
and
q, r
∈
K
[
x
]. But
f
(
θ
) =
q
(
θ
)
p
(
θ
) +
r
(
θ
)
Thus
r
(
θ
) = 0. If
r
is not identically zero, then
p
would not have minimal degree and so
r
= 0. Then
p
=
f
in
K
[
x
].
Theorem 2.
Let
K
⊆
C
.
If
f
∈
K
[
x
]
is irreducible in
K
[
x
]
of degree
n
, then
f
has
n
distinct roots in
C
.
Proof:
Suppose that in
C
[
x
],
f
(
x
) =
a
n
(
x

α
)
2
f
1
(
x
)
with
a
n
∈
C
,
α
∈
C
,
f
1
∈
C
[
x
]. Then
f
0
(
x
) = 2
a
n
(
x

α
)
f
1
(
x
) +
a
n
(
x

α
)
2
f
0
1
(
x
)
In particular,
f
0
(
α
) = 0.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '00
 NoneListed
 Number Theory, α, θ, Algebraic number theory

Click to edit the document details