Math 302
Modern Algebra
Spring 2009
Homework Solutions
3.1
DEFINITIONS AND EXAMPLES OF RINGS
2.
Let
R
=
{
0
, e, b, c
}
with addition and multiplication defined by the following tables. It can be shown
that
R
is a ring with this addition and multiplication.
+
0
e
b
c
0
0
e
b
c
e
e
0
c
b
b
b
c
0
e
c
c
b
e
0
·
0
e
b
c
0
0
0
0
0
e
0
e
b
c
b
0
b
b
0
c
0
c
0
c
(a)
Verify axioms R4 and R5.
R4. Existence of Additive Identity.
We need to show there exists an element 0
R
∈
R
such that
for all
a
∈
R
,
a
+ 0
R
=
a.
Proof:
Choose 0
R
= 0. Then 0
R
∈
R
. We need to show
(
∀
a
∈
R
) [
a
+ 0
R
=
a
]
.
To show this, we can use a proof by exhaustion. From the table,
0 + 0 = 0
,
0 +
e
=
e
=
e
+ 0
,
0 +
b
=
b
=
b
+ 0
,
0 +
c
=
c
=
c
+ 0
.
Therefore
a
+ 0 =
a
for all
a
∈
R
so 0 is an additive identity for
R
. Thus
R
satisfies R4.
R5. Existence of Additive Inverses.
We need to show for each
a
∈
R
, there exists
x
∈
R
such
that
a
+
x
= 0
R
.
Proof:
Use a proof by exhaustion. From the addition table, we have
0 + 0 = 0
,
e
+
e
= 0
,
b
+
b
= 0
,
c
+
c
= 0
,
which implies that each element is its own additive inverse. Therefore
R
satisfies R5.
(b)
Is
R
commutative?
R9. Commutative Law of Multiplication.
We need to show for all
x, y
∈
R
,
x
·
y
=
y
·
x.
Proof:
To check this, we again use a proof by exhaustion. We need to check all possible pairs
of (distinct) elements from
R
:
0
·
e
= 0 =
e
·
0
0
·
b
= 0 =
b
·
0
0
·
c
=
c
=
c
·
0
e
·
b
=
b
=
b
·
e
e
·
c
=
b
=
c
·
e
b
·
c
= 0 =
c
·
b
Therefore
R
satisfies R9 and so is a commutative ring.
23
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3.1 Definitions and Examples of Rings
Math 302 Homework Solutions page 24
2.
(continued)
(c)
Is
R
a field?
Since
R
is a commutative ring, to determine whether or not
R
is a field, we need to check
axioms R10 and R12.
R10. Existence of Multiplicative Identity.
There exists an element 1
R
∈
R
such that for all
a
∈
R
,
a
·
1
R
=
a
= 1
R
·
a.
Proof:
This property is satisfied. Choose 1
R
=
e
. Then 1
R
∈
R
. We need to show
(
∀
a
∈
R
) [
a
·
1
R
=
a
∧
1
R
·
a
=
a
]
.
To show this, we can use a proof by exhaustion. From the table,
0
·
e
= 0 =
e
·
0
,
e
·
e
=
e,
b
·
e
=
b
=
e
·
b,
c
·
e
=
c
=
e
·
c.
Therefore
e
is a multiplicative identity for
R
, so
R
satisfies R10 and
R
is a ring with identity.
R12. Existence of Multiplicative Inverses.
For each
a
∈
R
,
a
6
= 0
R
= 0, there exists
x
∈
R
such
that
a
·
x
= 1
R
=
e
.
Proof:
This property is not satisfied. For both
b
and
c
, there does not exist an element in
R
such that
b
·
x
=
e
and
c
·
x
=
e
. So neither of these elements has a multiplicative inverse.
Therefore
R
is not a field.
13.
Write out the addition and multiplication tables for
(a)
Z
2
×
Z
3
.
This is the Cartesian product of
Z
2
and
Z
3
, so
Z
2
×
Z
3
=
{
(
a, b
)

a
∈
Z
2
∧
b
∈
Z
3
}
=
{
(0
,
0)
,
(0
,
1)
,
(0
,
2)
,
(1
,
0)
,
(1
,
1)
,
(1
,
2)
}
,
with addition and multiplication defined as follows. For each (
a, b
)
,
(
c, d
)
∈
Z
2
×
Z
3
,
Addition:
(
a, b
) + (
c, d
) = (
a
+
c, b
+
d
)
Multiplication:
(
a, b
)
·
(
c, d
) = (
ac, bd
)
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 Spring '03
 Edwards
 Algebra, Addition, Multiplication, Multiplicative inverse, additive inverse, multiplicative inverses, R10

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