Exam 3 Solutions
1. Since 0
∈
A
and
ϕ
is a homomorphism,
ϕ
(0) = 0
∈
ϕ
(
A
) thus
ϕ
(
A
)
6
= Ø.
Let
ϕ
(
a
)
, ϕ
(
b
)
∈
ϕ
(
A
) and hence
a, b
∈
A
.
ϕ
(
a
)

ϕ
(
b
) =
ϕ
(
a

b
) because
ϕ
is a homomor
phism.
a

b
∈
A
because
A
is an ideal, hence
ϕ
(
a
)

ϕ
(
b
)
∈
ϕ
(
A
) and
ϕ
(
A
) is an additive
subgroup.
Let
s
∈
S
and
ϕ
(
a
)
∈
ϕ
(
A
). Since
ϕ
is onto, there exists
r
∈
R
such that
ϕ
(
r
) =
s
.
A
is
an ideal, so
ra
,
ar
∈
A
. Thus
sϕ
(
a
) =
ϕ
(
r
)
ϕ
(
a
) =
ϕ
(
ra
)
∈
ϕ
(
A
) and
ϕ
(
a
)
s
=
ϕ
(
a
)
ϕ
(
r
) =
ϕ
(
ar
)
∈
ϕ
(
A
).
Therefore
ϕ
(
A
) is an ideal of
S
.
2.
A
=
{
(3
x, y
)

x, y
∈
Z
}
is a maximal ideal.
•
Ideal
: note that for (3
a, b
)
,
(3
c, d
)
∈
A
that (3
a, b
)

(3
c, d
) = (3(
a

c
)
, b

d
)
∈
A
and
for (
a, b
)
∈
Z
×
Z
and (3
c, d
)
∈
A
(
a, b
)(3
c, d
) = (3
c, d
)(
a, b
) = (3
ac, bd
)
∈
A
(and clearly
A
is nonempty since (0
,
0) is in
A
). Thus
A
is an ideal.
•
Maximal
: Suppose
A
(
B
and
B
is an ideal of
Z
×
Z
. Then there exists (
b
1
, b
2
)
∈
B
such that (3
, b
1
) = 1.
Therefore there exists
α, β
∈
Z
such that 3
α
+
b
1
β
= 1.
Thus
(1
,
1) = (3
α
+
b
1
β,
1

b
2
+
b
2
) = (3
α,
1

b
2
)+(
b
1
β, b
2
) = (3
α,
1

b
2
)+(
β,
1)(
b
1
, b
2
)
∈
B
.
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 Spring '11
 Johnson
 Algebra, Prime number, Commutative ring, Principal ideal domain

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