Math 113 Homework # 8 solutions
Fraleigh 19.14. If
A
and
B
are matrices such that the image of
B
is contained
in the kernel of
A
, then
AB
= 0. This observation allows us to ﬁnd
that
±
1 2
2 4
²±
2
2

1

1
²
=
±
0 0
0 0
²
.
Fraleigh 19.17. (a) False,
n
Z
is a subring of
Z
which has no zero divisors.
(b) True, proved in class and the book.
(c) False,
n
Z
has characteristic zero since it is a subring of
Z
.
(d) False,
n
Z
has no multiplicative identity for
n >
1.
(e) True, because if a ring is isomorphic to an integral domain then it
is an integral domain.
(f) True. If it were ﬁnite then
m
·
1 =
n
·
1 for some
m
6
=
n
, so
(
m

n
)
·
1 = 0, so the characteristic divides
m

n
.
(g) False: (1
,
0)
·
(0
,
1) = (0
,
0).
(h) True. Suppose
a
is a zero divisor and has a multiplicative inverse.
Then
a
6
= 0, there exists
b
6
= 0 with
ab
= 0, and there exists
c
with
ac
= 1. Then 0 = 0
c
= (
ab
)
c
= (
ac
)
b
= 1
b
=
b
, contradiction.
(i) False,
n
Z
is not a domain when
n >
1 because it has no multiplica
tive identity.
(j) False,
Z
is not a ﬁeld.
Fraleigh 21.2.
F
=
{
a
+
b
√
2

a,b
∈
Q
}
. By a previous problem,
F
is a ﬁeld,
and
F
contains
D
. On the other hand every element of
F
is a quotient
of two elements of
D
, because
p/q
+ (
r/s
)
√
2 = (
ps
+
qr
√
2)
/
(
qs
).
Fraleigh 21.4. (a) True; this is essentially the deﬁnition of
Q
.
(b) False. If
R
were a ﬁeld of quotients of
Z
, then there would be an
injective ring homomorphism
f
:
Z
→
R
, such that every element of
R
is the quotient of two elements of the image of
f
. As explained in class,
an injective ring homomorphism