Proof.
Let
J
be the product of all the nonzero prime ideals. If
I
is any nonzero ideal,
then by (7.8.2) there is a nonzero ideal
I
such that
II
is a principal ideal (
a
), with
I
relatively prime to
J
. But then the set of prime factors of
I
is empty, so that
I
=
R
.
Thus (
a
) =
II
=
IR
=
I
.
♣
The next result shows that a Dedekind domain is not too far away from a principal
ideal domain.

7.9. P-ADIC NUMBERS
29
7.8.4
Corollary
Let
I
be a nonzero ideal of the Dedekind domain
R
, and let
a
be any nonzero element
of
I
. Then
I
can be generated by two elements, one of which is
a
.
Proof.
Since
a
∈
I
, we have (
a
)
⊆
I
, so
I
divides (
a
), say (
a
) =
IJ
. By (7.8.2) there is
a nonzero ideal
I
such that
II
is a principal ideal (
b
) and
I
is relatively prime to
J
. If
gcd stands for greatest common divisor, then the ideal generated by
a
and
b
is
gcd((
a
)
,
(
b
)) = gcd(
IJ,II
) =
I
since gcd(
J,I
) = (1).
♣
Problems For Section 7.8
1. Let
I
(
R
) be the group of nonzero fractional ideals of a Dedekind domain
R
. If
P
(
R
) is
the subset of
I
(
R
) consisting of all nonzero
principal fractional ideals
Rx
,
x
∈
K
, show
that
P
(
R
) is a subgroup of
I
(
R
). The quotient group
C
(
R
) =
I
(
R
)
/P
(
R
) is called the
ideal class group
of
R
. Since
R
is commutative,
C
(
R
) is abelian, and it can be shown
that in the number field case,
C
(
R
) is finite.
2. Continuing Problem 1, show that
C
(
R
) is trivial iff
R
is a PID.
We will now go through the factorization of an ideal in a number field. The neces-
sary background is developed in a course in algebraic number theory, but some of the
manipulations are accessible to us now. By (7.2.3), the ring
B
of algebraic integers of the
number field
Q
(
√
−
5) is
Z
[
√
−
5]. (Note that
−
5
≡
3 mod 4.) If we wish to factor the
ideal (2) = 2
B
in
B
, the idea is to factor
x
2
+5 mod 2, and the result is
x
2
+5
≡
(
x
+1)
2
mod 2. Identifying
x
with
√
−
5, we form the ideal
P
2
= (2
,
1+
√
−
5), which turns out to
be prime. The desired factorization is (2) =
P
2
2
. This technique works if
B
=
Z
[
α
], where
the number field
L
is
Q
(
α
).
3. Show that 1
−
√
−
5
∈
P
2
, and conclude that 6
∈
P
2
2
.
4. Show that 2
∈
P
2
2
, hence (2)
⊆
P
2
2
.
5. Expand
P
2
2
= (2
,
1 +
√
−
5)(2
,
1 +
√
−
5), and conclude that
P
2
2
⊆
(2).
6. Following the technique suggested in the above problems, factor
x
2
+ 5 mod 3, and
conjecture that the prime factorization of (3) in the ring of integers of
Q
(
√
−
5) is
(3) =
P
3
P
3
for appropriate
P
3
and
P
3
.
7. With
P
3
and
P
3
as found in Problem 6, verify that (3) =
P
3
P
3
.
7.9
p-adic Numbers
We will give a very informal introduction to this basic area of number theory. ( For further
details, see Gouvea, “p-adic Numbers”.) Throughout the discussion,
p
is a fixed prime.

30
CHAPTER 7. INTRODUCING ALGEBRAIC NUMBER THEORY
7.9.1
Definitions and Comments
A
p
-adic integer
can be described in several ways. One representation is via a series
x
=
a
0
+
a
1
p
+
a
2
p
2
+
· · ·
, a
i
∈
Z
.
(1)
(Let’s ignore the problem of convergence for now.) The partial sums are
x
n
=
a
0
+
a
1
p
+
· · ·
+
a
n
p
n
, so that
x
n
−
x
n
−
1
=
a
n
p
n
. A
p
-adic integer can also be defined as a sequence
of integers
x
=
{
x
0
,x
1
,...,
}
satisfying
x
n
≡
x
n
−
1
mod
p
n
, n
= 1
,
2
,
....
(2)
Given a sequence satisfying (2), we can recover the coeﬃcients of the series (1) by
a
0
=
x
0
, a
1
=
x
1
−
x
0
p
, a
2

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