110
some constant
c
. If
x
is a multiple of
ax
+
by
+
c
, then
b
=
0; if
y
is a mul
tiple, then
a
=
0. We conclude that
d
(
x
,
y
)
is a nonzero constant. Since
k
is a
fi
eld,
d
is a unit, and so
(
x
,
y
)
=
(
d
)
=
k
[
x
,
y
]
, a contradiction.
3.76
For every
m
≥
1, prove that every ideal in
I
m
is a principal ideal. (If
m
is
composite, then
I
m
is not a PID because it is not a domain.)
Solution.
Let
I
be an ideal in
I
m
.
Consider the ring homomorphism
ν
:
Z
→
I
m
de
fi
ned by
ν
:
n
→ [
n
]
; of course,
ν
is a surjection.
It is
easily seen that the inverse image
J
=
ν
−
1
(
I
)
= {
n
∈
Z
:
ν(
n
)
∈
I
}
is an ideal in
Z
; moreover,
ν(
J
)
=
I
. But every ideal in
Z
is principal, so
that
J
=
(
a
)
. It follows that
I
=
(
[
a
]
)
.
This argument really shows that if
R
is a PID and
L
is an ideal in
R
, then
every ideal in the quotient ring
R
/
L
is principal.
3.77
Let
R
be a PID and let
π
∈
R
be an irreducible element. If
β
∈
R
and
π
β
, prove that
π
and
β
are relatively prime.
Solution.
3.78
(i)
Show that
x
,
y
∈
k
[
x
,
y
]
are relatively prime but that 1 is not a lin
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 Fall '11
 KeithCornell
 Prime number, Integral domain, Ring theory

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