proof of characterizations of the Jacobson radical

proof of characterizations of the Jacobson radical

First, note that by

definition

a left

primitive ideal

is the

annihilator

of an

irreducible

left

R

-module, so clearly

characterization

1) is

equivalent

to the definition of the

Jacobson radical

.

Next, we will prove cyclical containment. Observe that 5) follows after the equivalence of 1) - 4) is established, since 4) is independent of the choice of left or

right ideals

.

1)

⊂

2)

We know that every left primitive ideal is the
largest ideal

contained

in a

maximal

left ideal

. So the

intersection

of all left primitive ideals will be contained in the intersection of all
maximal left ideals.

2)

⊂

3)

Let

S

=

{

M

:

M

⁢

a maximal left ideal of

⁢

R

}

and take

r

∈

R

.
Let

t

∈

∩

M

∈

S

M

. Then

r

⁢

t

∈

∩

M

∈

S

M

.

Assume

1

-

r

⁢

t

is not

left invertible

; therefore there exists a maximal
left ideal

M

0

of

R

such that

R

⁢

(

1

-

r

⁢

t

)

⊆

M

0

.

Note then that

1

-

r

⁢

t

∈

M

0

. Also, by definition of

t

, we have

r

⁢

t

∈

M

0

. Therefore

1

∈

M

0

; this

contradiction

implies

1

-

r

⁢

t

is
left invertible.

3)

⊂

4)

We claim that 3)

satisfies

the condition of 4).

Let

K

=

{

t

∈

R

:

1

-

r

⁢

t

⁢

is left invertible for all

⁢

r

∈

R

}

.

We shall first show that

K

is an ideal.

Clearly if

t

∈

K

, then

r

⁢

t

∈

K

.
If

t

1

,

t

2

∈

K

, then

1

-

r

⁢

(

t

1

+

t

2

)

=

(

1

-

r

⁢

t

1

)

-

r

⁢

t

2

Now there exists

u

1

such that

u

1

⁢

(

1

-

r

⁢

t

1

)

=

1

, hence

u

1

⁢

(

(

1

-

r

⁢

t

1

)

-

r

⁢

t

2

)

=

1

-

u

1

⁢

r

⁢

t

2

Similarly, there exists

u

2

such that

u

2

⁢

(

1

-

u

1

⁢

r

⁢

t

2

)

=

1

, therefore

u

2

⁢

u

1

⁢

(

1

-

r

⁢

(

t

1

+

t

2

)

)

=

1

Hence

t

1

+

t

2

∈

K

.

Now if

t

∈

K

,

r

∈

R

, to show that

t

⁢

r

∈

K

it suffices to
show that

1

-

t

⁢

r

is left invertible. Suppose

u

⁢

(

1

-

r

⁢

t

)

=

1

, hence

u

-

u

⁢

r

⁢

t

=

1

, then

t

⁢

u

⁢

r

-

t

⁢

u

⁢

r

⁢

t

⁢

r

=

t

⁢

r

.

So

(

1

+

t

⁢

u

⁢

r

)

⁢

(

1

-

t

⁢

r

)

=

1

+

t

⁢

u

⁢

r

-

t

⁢

r

-

t

⁢

u

⁢

r

⁢

t

⁢

r

=

1

.

Therefore

K

is an ideal.

Now let

v

∈

K

. Then there exists

u

such that

u

⁢

(

1

-

v

)

=

1

, hence

1

-

u

=

-

u

⁢

v

∈

K

, so

u

=

1

-

(

1

-

u

)

is left invertible.

So there exists

w

such that

w

⁢

u

=

1

, hence

w

⁢

u

⁢

(

1

-

v

)

=

w

, then

1

-

v

=

w

.
Thus

(

1

-

v

)

⁢

u

=

1

and therefore

1

-

v

is a unit.

Let

J

be the largest ideal such that, for all

v

∈

J

,

1

-

v

is a
unit. We claim that

K

⊆

J

.

Suppose this were not true; in this case

K

+

J

strictly

contains

J

.
Consider

r

⁢

x

+

s

⁢

y

∈

K

+

J

with

x

∈

K

,

y

∈

J

and

r

,

s

∈

R

.
Now

1

-

(

r

⁢

x

+

s

⁢

y

)

=

(

1

-

r

⁢

x

)

-

s

⁢

y

, and since

r

⁢

x

∈

K

, then

1

-

r

⁢

x

=

u

for some
unit

u

∈

R

.

So

1

-

(

r

⁢

x

+

s

⁢

y

)

=

u

-

s

⁢

y

=

u

⁢

(

1

-

u

-

1

⁢

s

⁢

y

)

, and clearly

u

-

1

⁢

s

⁢

y

∈

J

since

y

∈

J

. Hence

1

-

u

-

1

⁢

s

⁢

y

is also a unit, and thus

1

-

(

r

⁢

x

+

s

⁢

y

)

is
a unit.

Thus

1

-

v

is a unit for all

v

∈

K

+

J

. But this contradicts the

assumption

that

J

is the largest such ideal. So we must have

K

⊆

J

.

4)

⊂

1)

We must show that if

I

is an ideal such that
for all

u

∈

I

,

1

-

u

is a unit, then

I

⊂

ann

⁡

(

M

R

)

for every irreducible left

R

-module

M

R

.

Suppose this is not the case, so there exists

M

R

such that

I

⊄

ann

⁡

(

M

R

)

. Now we know that

ann

⁡

(

M

R

)

is the largest ideal inside some maximal
left ideal

J

of

R

. Thus we must also have

I

⊄

J

,
or else this would contradict the maximality of

ann

⁡

(

M

R

)

inside

J

.

But since

I

⊄

J

, then by maximality

I

+

J

=

R

, hence there
exist

u

∈

I

and

v

∈

J

such that

u

+

v

=

1

. Then

v

=

1

-

u

, so

v

is a unit and

J

=

R

. But since

J

is a proper left ideal, this
is a contradiction.

Title

proof of characterizations of the Jacobson radical

Canonical

name

ProofOfCharacterizationsOfTheJacobsonRadical

Date of creation

2013-03-22 12:48:56

Last modified on

2013-03-22 12:48:56

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

31

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 16N20

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