Math 567 Number Theory I
Homework 8
Name: Zhaoning Yang
November 3, 2013
Problem 1
(Textbook P.234 #9.13) Let
K
be an algebraic number field and
O
K
be its ring of integers.
Let
m
∈
Z
\{
0
}
. Prove that there exists only finitely many integral ideals of
O
K
to which
m
belongs.
Solution:
Proof.
Let
K
be an algebraic number field. Let
I
be an integral ideal of
O
K
. SInce
O
K
is a Dedekind
domain,
m
∈
I
⇐⇒ h
m
i ⊆
I
⇐⇒
I
h
m
i
. So it suffice to prove there are only finitely many ideals which
divides
h
m
i
. Because
m
∈
Z
\{
0
}
, m
∈ O
K
\{
0
}
. It follows that
h
m
i
is a nonzero integral ideal in
O
K
.
Then it has a unique factorization as products of prime ideals
h
m
i
=
n
Y
i
=1
p
a
i
i
where
a
i
∈
Z
+
,
p
i
are distinct prime ideals of
O
K
for all 1
≤
i
≤
n
. So if
I
is a ideal that divides
h
m
i
,
from above factorization
I
=
Q
n
i
=1
p
b
i
i
where 0
≤
b
i
≤
a
i
, b
i
∈
Z
for all 1
≤
i
≤
n
with
b
i
s not all zero.
Since for each
b
i
there are only finitely many choices, we can say
I
=
Q
n
i
=1
p
b
i
i
also has finitely many
choices. Therefore there are only finitely many integral ideals that divides
h
m
i
.
Problem 2
(Textbook P.234 #9.17) Let
K
be an algebraic number field. Let
O
K
be its ring of integers
and
p
∈
Spec
O
K
. Let
a
∈ O
K
be such that
p
does not divide
h
a
i
. Prove that
a
N
(
p
)

1
≡
1(mod
p
).
Solution:
Proof.
Since
p
is a prime ideal in
O
K
. We know
O
K
/
p
is an integral domain. Moreover, since
O
K
is a
Dedekind domain,
p
is also a maximal ideal of
O
K
, hence
O
K
/
p
is a field. From now on, we use the notation
b
denote the coset representative
b
+
p
in
O
K
/
p
. Let
G
= (
O
K
/
p
)
*
=
O
K
/
p
 {
0
}
be the multiplicative
group of
O
K
/
p
. Since
p
does not divide
h
a
i
, we know
a /
∈
p
for
a
∈
p
⇐⇒ h
a
i ⊆
p
⇐⇒
p
h
a
i
. Therefore
a
=
a
+
p
∈ O
K
/
p
 {
0
}
= (
O
K
/
p
)
*
=
G
. From
Theorem 9.1.3
since
p
is a nonzero ideal in
O
K
, we
know
N
(
p
) = [
O
K
:
p
]
∈
Z
+
. It follows that

G

=
O
K
/
p
 
1 = [
O
K
:
p
]

1 =
N
(
p
)

1
Since
a
∈
G
, it follows from basic group theory that
a

G

=
id
G
. (i.e. in a finite group, the order of any
elements must divide the order of the group since by Lagrange’s Theorem

G

= [
G
:
H
]
· 
H

for any
H
subgroup of
G
) Because
id
G
= 1 +
p
∈ O
K
/
p
, and
a

G

= (
a
+
p
)

G

=
a

G

+
p
=
a
N
(
p
)

1
+
p
, we can say
a

G

=
id
G
⇐⇒
a
N
(
p
)

1
+
p
= 1 +
p
⇐⇒
a
N
(
p
)

1

1
∈
p
⇐⇒
a
N
(
p
)

1
≡
1( mod
p
)
So we have shown
a
N
(
p
)

1
≡
1(mod
p
) as expected.
1
Problem 3
(Textbook P.234 #9.19) Determine all complex quadratic fields
K
for which
O
K
possesses
elements of norm 38 and trace 11.
Solution:
We want to find
m
∈
Z
square free such that when
K
=
Q
(
√
m
),
O
K
has elements of norm 38
and trace 11. First, note this
m
must congruent 1 modulo 4, for if
m
6
= 1(mod 4), then
O
K
=
Z
+
Z
√
m
.
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 YANG
 Algebra, Number Theory, Integers, Prime number, Algebraic number theory, Integral domain, Principal ideal domain, prime ideal