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88
On the other hand,
a
◦
(
b
◦
c
)
=
a
◦
(
b
+
c
−
bc
)
=
a
+
b
+
c
−
bc
−
a
(
b
+
c
−
bc
)
=
a
+
b
+
c
−
bc
−
ab
−
ac
+
a
(
bc
).
(ii)
Prove that a commutative ring
R
is a
f
eld if and only if the set
R
#
={
r
∈
R
:
r
±=
1
}
is an abelian group under the circle operation.
Solution.
Assume that
R
is a
f
eld. To show that
R
#
is a group,
we must
f
rst show that
◦
is an operation on
R
#
; that is, if
a
±=
1
and
c
±=
1, then
a
◦
c
±=
1. If, on the contrary,
1
=
a
◦
c
=
a
+
c
−
=
a
+
c
(
1
−
a
),
then 1
−
a
=
c
(
1
−
a
)
. Since
a
±=
1, we may cancel 1
−
a
,
and this gives the contradiction
c
=
1. Part (i) shows that
◦
is
associative and that 0 acts as the identity. It remains to
f
nd the
inverse of an element
a
±=
1; that is, to
f
nd
b
±=
1 such that
0
=
a
◦
b
=
a
+
b
−
.De
f
ne
b
=
a
(
a
−
1
)
−
1
.
Conversely, assume that
R
#
is a group. It suf
f
ces to
f
nd a (multi
plicative) inverse for every
a
±=
0. Now
a
±=
0 implies
a
+
1
±=
1,
and so there exists
b
∈
R
#
with
(
a
+
1
)
◦
b
=
0; that is,
0
=
a
+
1
+
b
−
(
a
+
1
)
b
=
a
+
1
+
b
−
−
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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