101
3.47
Let
ψ
:
R
→
S
be a homomorphism, where
R
and
S
are commutative
rings, and let ker
ψ
=
I
.If
J
is an ideal in
S
, prove that
ψ
−
1
(
J
)
is an ideal
in
R
which contains
I
.
Solution.
Absent.
3.48
If
R
is a commutative ring and
c
∈
R
, prove that the function
ϕ
:
R
[
x
]→
R
[
x
]
,de
f
ned by
f
(
x
)
7→
f
(
x
+
c
)
, is an isomorphism. In more detail,
ϕ(
∑
i
s
i
x
i
)
=
∑
i
s
i
(
x
+
c
)
i
.
Solution.
It is clear that
ϕ(
1
)
=
1. If
f
(
x
)
=
∑
s
j
x
j
and
g
(
x
)
=
∑
r
j
x
j
,
then
ϕ(
f
+
g
)
=
ϕ(
X
(
s
i
+
r
i
)
x
i
)
=
X
(
s
i
+
r
i
)(
x
+
c
)
i
=
X
s
i
(
x
+
c
)
i
+
X
r
i
(
x
+
c
)
i
=
ϕ(
f
)
+
ϕ(
g
),
as desired.
To prove that
ϕ(
fg
)
=
(ϕ
f
)(ϕ
g
)
,
f
rst consider the special case
f
(
x
)
=
sx
k
, where
s
∈
R
and
k
≥
0. If
g
(
x
)
=
∑
r
i
x
i
, then
ϕ(
sx
k
f
(
x
))
=
ϕ(
X
sr
i
x
i
+
k
)
=
X
sr
i
(
x
+
c
)
i
+
k
=
s
(
x
+
c
)
k
X
r
i
(
x
+
c
)
i
=
ϕ(
sx
k
)ϕ(
g
).
We now prove that
ϕ(
fg
)
=
ϕ(
f
)ϕ(
g
)
by induction on deg
(
f
)
. The base
step deg
(
f
)
=
0 is the special case just proved when
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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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