Math 4124
Wednesday, May 2
Tenth Homework Solutions
1. (a) Let
s
,
t
∈
S
and suppose
st
/
∈
S
. Then
stx
=
0 for some nonzero
x
∈
R
. Then
tx
6
=
0
because
t
is a nonzero divisor, and hence
stx
because
s
is a nonzero divisor and
tx
6
=
0. It follows that
S
is a multiplicatively closed subset of
R
. Obviously it
contains 1 but not 0.
(b) Let
r
/
s
∈
S

1
R
, where
r
∈
R
and
s
∈
S
. If
r
is a zero divisor, then
rx
=
0 for some
nonzero
x
∈
R
, hence
(
r
/
s
)(
x
/
1
) =
0 and
x
/
1
6
=
0. This shows that
r
/
s
is a zero
divisor. On the other hand if
r
is not a zero divisor, then
r
∈
S
by deﬁnition of
S
and thus
s
/
r
∈
S

1
R
. Since
(
r
/
s
)(
s
/
r
) =
1, we see that
r
/
s
is a unit and we’re
ﬁnished.
2. Let
K
=
ker
α
and
J
=
ker
β
. Then
R
/
K
∼
=
Z
/
2
Z
, a ﬁeld, and we see that
K
is a
maximal ideal of
R
. Similarly
J
is a maximal ideal of
R
. Since

R
/
J
 6
=

R
/
K

, we see
that
J
6
=
K
and hence
J
+
K
=
R
. By the Chinese Remainder theorem, we see that
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 Spring '08
 Staff
 Math, Algebra, Division, Prime number, Divisor, Integral domain, prime ideal, nonzero divisor

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