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6.3. QUOTIENT RINGS
285
.a
C
bx/
; evaluating at
i
, we get
0
D
a
C
bi
, which is possible only if
a
D
b
D
0
. Thus
g
is a multiple of
x
2
C
1
. That is ker
.'/
D
.x
2
C
1/
.
By the homomorphism theorem for rings,
R
ŒxŁ=.x
2
C
1/
Š
C
as rings.
In particular, since
C
is a ﬁeld,
R
ŒxŁ=.x
2
C
1/
is a ﬁeld. Note that we
have already calculated explicitly in Example
6.3.3
that multiplication in
R
ŒxŁ=.x
2
C
1/
satisﬁes the same rule as multiplication in
C
.
Example 6.3.6.
Let
R
be a ring with identity containing ideals
B
1
;
:::;B
s
. Let
B
D \
i
B
i
. Suppose that
B
i
C
B
j
D
R
for all
i
¤
j
. Then
R=B
Š
R=B
1
˚ ±±± ˚
R=B
s
. In fact,
'
W
r
7!
.r
C
B
1
;:::;r
C
B
s
/
is a
homomorphism of
R
into
R=B
1
˚±±±˚
R=B
s
with kernel
B
, so
R=B
Š
'.R/
. The problem is to show that
'
is surjective. Fix
i
and for each
j
¤
i
ﬁnd
r
0
j
2
B
i
and
r
j
2
B
j
such that
r
0
j
C
r
j
D
1
. Consider the product
of all the
.r
0
j
C
r
j
/
(in any order). When the product is expanded, all the
summands except for one contain at least one factor
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 Fall '08
 EVERAGE
 Algebra

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