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College of the Holy Cross, Fall 2009
Math 351 – Homework 3 selected solutions
Chapter 14, #28
First one must verify that
A
=
{
(3
x,y
)

x,y
∈
Z
}
, the set of all elements whose
ﬁrst entry is a multiple of 3 and whose second entry is allowed to be anything, is in fact an ideal
of
Z
⊕
Z
. This is straightforward to do using the ideal test, and everyone basically got it. To show
that
A
is maximal, let
B
be an ideal of
Z
⊕
Z
, and assume that
B
properly contains
A
(that is,
contains
A
but is not equal to
A
). Then
B
contains at least one element not in
A
. Say this element
is (
r,s
). Since (
r,s
)
6∈
A
, we know that
r
cannot be a multiple of 3. Say that
r
= 3
n
+ 1 for some
n
∈
Z
(the other case is
r
= 3
n
+ 2 for some
n
∈
Z
). Then (3
n,s

1)
∈
A
, and so (3
n,s

1)
∈
B
.
Thus since
B
is closed under subtraction, we get (3
n
+ 1
,s
)

(3
n,s

1) = (1
,
1)
∈
B
. However,
(1
,
1) is the multiplicative identity in
Z
⊕
Z
, and it follows from remarks in class that
B
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