commutativity theorems on rings

commutativity theorems on rings

Since Wedderburn proved his celebrated

theorem

that any finite

division ring

is

commutative

, the interest in studying properties on a ring that would render the ring commutative dramatically increased. Below is a list of some of the so-called “commutativity theorems” on a ring, showing how much one can generalize the result that Wedderburn first obtained. In the list below,

R

is assumed to be

unital ring

.

Theorem 1

.

In each of the cases below,

R

is commutative:

1.

(Wedderburn’s theorem)

R

is a finite division ring.

2.

(Jacobson) If for every

element

a

∈

R

, there is a positive integer

n

>

1

(depending on

a

), such that

a

n

=

a

.

3.

(Jacobson-Herstein) For every

a

,

b

∈

R

, if there is a positive integer

n

>

1

(depending on

a

,

b

) such that

(

a

⁢

b

-

b

⁢

a

)

n

=

a

⁢

b

-

b

⁢

a

.

4.

(Herstein) If there is an integer

n

>

1

such that for every element

a

∈

R

such that

a

n

-

a

∈

Z

⁢

(

R

)

, the center of

R

.

5.

(Herstein) If for every

a

∈

R

, there is a

polynomial

p

∈

ℤ

⁢

[

X

]

(

p

depending on

a

) such that

a

2

⁢

p

⁢

(

a

)

-

a

∈

Z

⁢

(

R

)

.

6.

(Herstein) If for every

a

,

b

∈

R

, such that there is an integer

n

>

1

(depending on

a

,

b

) with

(

a

n

-

a

)

⁢

b

=

b

⁢

(

a

n

-

a

)

.

Some of the commutativity problems can be derived fairly easily, such as the following examples:

Theorem 2

.

If

R

is a ring with

1

such that

(

a

⁢

b

)

2

=

a

2

⁢

b

2

for all

a

,

b

∈

R

, then

R

is commutative.

Proof.

Let

a

,

b

∈

R

. From the

assumption

, we have

(

(

a

+

1

)

⁢

b

)

2

=

(

a

+

1

)

2

⁢

b

2

. Expanding the LHS, we get

(

a

⁢

b

)

2

+

(

a

⁢

b

)

⁢

b

+

b

⁢

(

a

⁢

b

)

+

b

2

. Expanding the RHS, we get

a

2

⁢

b

2

+

2

⁢

a

⁢

b

2

+

b

2

. Equating both sides and eliminating common terms, we have

b

⁢

a

⁢

b

=

a

⁢

b

2

(1)

Similarly, from

(

a

⁢

(

b

+

1

)

)

2

=

a

2

⁢

(

b

+

1

)

2

, we expand the

equations

and get

(

a

⁢

b

)

2

+

(

a

⁢

b

)

⁢

a

+

a

⁢

(

a

⁢

b

)

+

a

2

=

a

2

⁢

b

2

+

2

⁢

a

2

⁢

b

+

a

2

.

Hence

a

⁢

b

⁢

a

=

a

2

⁢

b

(2)

Finally, expanding out

(

(

a

+

1

)

⁢

(

b

+

1

)

)

2

=

(

a

+

1

)

2

⁢

(

b

+

1

)

2

and eliminating
common terms, keeping in mind Equations (1) and (2) from above, we get

a

⁢

b

=

b

⁢

a

.
∎

Corollary 3

.

If each element of a ring

R

is

idempotent

, then

R

is commutative.

Proof.

If

R

contains

1

, then we can apply Theorem 2: for

(

s

⁢

t

)

2

=

s

⁢

t

=

s

2

⁢

t

2

for any

s

,

t

∈

R

. Otherwise, we do the following trick: first

2

⁢

s

=

(

2

⁢

s

)

2

=

4

⁢

s

2

=

4

⁢

s

, so that

2

⁢

s

=

0

for all

s

∈

R

. Next,

s

+

t

=

(

s

+

t

)

2

=

s

2

+

s

⁢

t

+

t

⁢

s

+

t

2

=

s

+

s

⁢

t

+

t

⁢

s

+

t

, so

0

=

s

⁢

t

+

t

⁢

s

, which implies

s

⁢

t

=

s

⁢

t

+

(

s

⁢

t

+

t

⁢

s

)

=

2

⁢

s

⁢

t

+

t

⁢

s

=

t

⁢

s

, and the result follows.

The corollary also follows directly from part 2 of Theorem 1.

∎

References

1

I. N. Herstein,

Noncommutative Rings

, The

Mathematical Association of America

(1968).

Title

commutativity theorems on rings

Canonical name

CommutativityTheoremsOnRings

Date of creation

2013-03-22 17:54:55

Last modified on

2013-03-22 17:54:55

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

12

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 16B99

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