Math 5125
Friday, October 28
Second Test Review
Sample Second Test no. 3. Answer All Problems.
Please Give Explanations For Your Answers.
1. Let
R
be a commutative ring which has nonprincipal ideals. Prove that
R
has a max
imal nonprincipal ideal (that is, an ideal
I
of
R
such that
I
is not principal, but if
I
⊂
J
±
R
, then
J
is principal).
(16 points)
2. Prove that 2, 3
±
√
13 are irreducible in
Z
[
√
13
]
. Also prove that 2 is not prime in
Z
[
√
13
]
.
(17 points)
3. Let
R
be a UFD and let
P
be a nonzero principal ideal of
R
.
(a) Prove that there are only ﬁnitely many principal ideals of
R
containing
P
.
(b) Is it true that there are only ﬁnitely many principal ideals of
R
/
P
? (Prove or give
counter example.)
(17 points)
Solution to Problem 2 on Sample Test no. 1
First note that
S

1
M
±
S

1
R
. Obviously 0
∈
S

1
M
. If
m
/
s
,
n
/
t
∈
S

1
M
, then
m
/
s
+
n
/
t
= (
mt
+
ns
)
/
(
st
)
∈
S

1
M
. Also if
r
/
u
∈
S

1
R
,
then
(
m
/
s
)(
r
/
u
) = (
mr
)
/
(
su
)
∈
S

1
M
.
If
S
∩
M
6
=
/0, choose
s
∈
S
∩
M
. Then
s
/
s
=
1
∈
S

1
M
, which shows that
S

1
M
=
S

1
R
and
hence
S

1
M
is not a maximal ideal.
Now suppose
S
∩
M
=
/0. If
S

1
M
=
S

1
R
, then 1
/
1
=
m
/
s
for some
m
∈
M
and
s
∈
S
.
This tells us that
(
m

s
)
t
=
0 for some
t
∈
S
and we deduce that
st
∈
M
which contradicts
S
∩
M
=
/0. Therefore
S

1
M
6
=
S

1
R
. Finally suppose
S

1
M
⊆
J
±
S

1
R
and
J
6
=
S

1
R
.
Let
θ
:
R
→
S

1
R
denote the natural homomorphism
r
7→
r
/
1. Then
M
⊆
θ

1
J
±
R
and
θ

1
J
6
=
R
. Therefore
θ

1
J
=
M
by maximality of
M
. If
x
∈
J
, then
x
=
y
/
s
for some
s
∈
S
and
y
∈
R
, and
y
/
1
∈
J
∩
θ
(
R
)
. Therefore
y
∈
θ

1
J
and we deduce that
y
∈
M
. Then
x
=
y
/
s
∈
S

1
M
, consequently
J
=
S

1
M
and we conclude that
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 Fall '07
 PALinnell
 Math, Algebra, Ring, Prime number, Integral domain, Commutative ring, Principal ideal domain

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