Math 7330: Functional Analysis
Fall 2002
Notes/Homework 4: Commutative Banach Algebras I
A. Sengupta
1. Let
R
be a commutative ring with multiplicative identity
e
. A subset
S
⊂
R
is an
ideal
of
R
if :(a) 0
∈
S
, (b)
x
+
y
∈
S
for every
x, y
∈
S
, and (c)
rx
∈
S
for every
r
∈
R
and
x
∈
S
. The ideal
S
is a
proper
ideal if
S
6
=
R
. It is a
maximal ideal
if it is
a proper ideal and if the only ideals containing
S
are
S
itself and the whole ring
R
.
The ideal
S
is a
prime
ideal if for every
x, y
∈
S
if
xy
∈
S
then at least one of
x
and
y
must be in
S
.
(i) Let
I
be an ideal of
R
. For any
x
∈
R
we write
x
+
I
be the set of all elements of
the form
x
+
i
with
i
running over
I
. Let
R/I
be the set of all sets of the form
x
+
I
with
x
running over
R
:
R/I
def
=
{
x
+
I
:
x
∈
R
}
Let
p
:
R
→
R/I
:
x
7→
x
+
I
For any elements
a, b
∈
R
we have
p
(
a
) =
p
(
b
) if and only if
a

b
∈
I
Show that if
x, x
0
, y, y
0
∈
R
are such that
p
(
x
) =
p
(
x
0
) and
p
(
y
) =
p
(
y
0
) then
p
(
x
+
x
0
) =
p
(
y
+
y
0
) and
p
(
xy
) =
p
(
yy
0
).
Thus there are welldefined operations of addition and multiplication on
R/I
given by
p
(
x
) +
p
(
y
)
def
=
p
(
x
+
y
)
,
p
(
x
)
p
(
y
)
def
=
p
(
xy
)
As is readily checked, these operations make
R/I
a ring and, of course,
p
:
R
→
R/I
is a ring homomorphism. Commutativity of
R
implies that
R/I
is commu
tative.
If
e
∈
R
is the identity of
R
then
p
(
e
) is the multiplicative identity in
R/I
.
1
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(ii) Suppose
I
is a maximal ideal of
R
. Show that then the commutative ring
R/I
is
a
field
, i.e. every nonzero element has an inverse. Hint: Let
x
∈
R
be such that
p
(
x
) is a nonzero element of
R/I
, i.e.
x
∈
R
is not in the ideal
I
. The set
Rx
+
I
=
{
rx
+
y
:
r
∈
R, y
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 Spring '08
 Davidson,M
 Algebra, Vector Space, Ring, Banach space

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