College of the Holy Cross, Fall 2009
Math 352, Practice Midterm 1 solutions
Prof. Jones
1. Let
I
be an ideal in a ring
R
, and set
N
(
I
) =
{
x
∈
R
:
xi
= 0 for all
i
∈
I
}
.
a) Show that
N
(
I
) is an ideal in
R
.
Using the ideal test, we must show ﬁrst that
N
(
I
) is not empty, and then that that if
x,y
∈
N
(
I
) and
r
∈
R
, then
x

y
,
rx
, and
xr
are all in
N
(
I
). Note that 0
∈
N
(
I
), proving that
N
(
I
) isn’t empty.
Now
x,y
∈
N
(
I
) means that
xi
=
yi
= 0 for all
i
∈
I
, and so (
x

y
)
i
=
xi

yi
= 0

0 = 0 for
all
i
∈
I
. Hence
x

y
∈
N
(
I
). Also, (
rx
)
i
=
rxi
=
r
0 = 0 for all
i
∈
I
, and thus
rx
∈
N
(
I
).
Consider (
xr
)
i
, which is the same as
x
(
ri
). Since
I
is an ideal,
ri
∈
I
. But
x
times any
element of
I
must be 0 (by the deﬁnition of
N
(
I
)), and thus
x
(
ri
) = 0.
b) Show that if
R
is an integral domain, then either
I
=
{
0
}
or
N
(
I
) =
{
0
}
.
Suppose that
I
6
=
{
0
}
, and suppose that
x
∈
N
(
I
). Then
xi
= 0 for all
i
∈
I
. Since
I
6
=
{
0
}
,
there is some
i
∈
I
with
i
= 0. Hence
xi
= 0 for some
i
6
= 0. Since
R
is an integral domain,
this implies that
x
= 0 (otherwise both
x
and
i
would be zerodivisors). We’ve thus taken an
arbitrary
x
∈
N
(
I
) and shown that
x
= 0. Thus
N
(
I
) =
{
0
}
.
c) Give an example of a ring
R
and an ideal
I
6
=
{
0
}
in
R
such that
N
(
I
) =
R
.
Consider the subring
R
=
{
0
,
2
}
of
Z
4
. Let
I
=
R
. Then for any
x
∈
R
, we have
xi
= 0 for
all
i
∈
I
. If
x
= 0 then this is obvious, and if
x
= 2 it is still true that 2
i