Math 3124
Thursday, November 10
November 10, Ungraded Homework
Problem 25.3 on page 127
Verify that
([
2
]
,
[
0
])
is a zero divisor in
Z
3
×
Z
3
.
Obviously
([
2
]
,
[
0
])
6
= ([
0
]
,
[
0
])
. Also
([
2
]
,
[
0
])([
0
]
,
[
2
]) = ([
0
]
,
[
0
])
. Since
([
0
]
,
[
2
])
6
=
0, this
shows that
([
2
]
,
[
0
])
is a zero divisor in
Z
3
×
Z
3
.
Problem 25.6 on page 127
What is the smallest subring of
R
containing 1/2?
Let
R
be the smallest subring of
R
containing 1/2. Then 1
/
2
+
1
/
2
=
1
∈
R
. Since
R
is
an abelian group under +, we deduce that
Z
⊆
R
. Also
R
is closed under multiplication,
consequently
a
/
2
b
∈
R
for all
a
∈
Z
and all nonnegative integers
b
. Now write
S
for this
subset, that is
S
=
{
a
/
2
b

a
,
b
∈
Z
,
b
≥
0
}
. Then
S
⊆
R
. All that remains to show is that
S
is a subring of
R
. First note that 0
∈
S
. Suppose
a
/
2
b
and
c
/
2
d
∈
S
. Then
a
/
2
b
+
c
/
2
d
=
(
2
d
a
+
2
b
c
)
/
2
b
+
d
and

(
a
/
2
b
) = (

a
)
/
2
b
, which shows that
S
is closed under + and taking
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 Fall '08
 PARRY
 Math, Algebra, Ring, Integral domain, Ring theory

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