proof of Wedderburn’s theorem

proof of Wedderburn’s theorem

We want to show that the

multiplication

operation

in a finite division
ring is

abelian

.

We denote the

centralizer

in

D

of an

element

x

as

C

D

⁢

(

x

)

.

Lemma. The centralizer is a subring.

0

and

1

are obviously

elements of

C

D

⁢

(

x

)

and if

y

and

z

are,
then

x

⁢

(

-

y

)

=

-

(

x

⁢

y

)

=

-

(

y

⁢

x

)

=

(

-

y

)

⁢

x

,

x

⁢

(

y

+

z

)

=

x

⁢

y

+

x

⁢

z

=

y

⁢

x

+

z

⁢

x

=

(

y

+

z

)

⁢

x

and

x

⁢

(

y

⁢

z

)

=

(

x

⁢

y

)

⁢

z

=

(

y

⁢

x

)

⁢

z

=

y

⁢

(

x

⁢

z

)

=

y

⁢

(

z

⁢

x

)

=

(

y

⁢

z

)

⁢

x

, so

-

y

,

y

+

z

, and

y

⁢

z

are also
elements of

C

D

⁢

(

x

)

. Moreover, for

y

≠

0

,

x

⁢

y

=

y

⁢

x

implies

y

-

1

⁢

x

=

x

⁢

y

-

1

, so

y

-

1

is also an element of

C

D

⁢

(

x

)

.

Now we consider the center of

D

which we’ll call

Z

⁢

(

D

)

. This is
also a subring and is in fact the

intersection

of all centralizers.

Z

⁢

(

D

)

=

⋂

x

∈

D

C

D

⁢

(

x

)

Z

⁢

(

D

)

is an abelian subring of

D

and is thus a field. We can consider

D

and every

C

D

⁢

(

x

)

as

vector spaces

over

Z

⁢

(

D

)

of dimension

n

and

n

x

respectively. Since

D

can be viewed as a module over

C

D

⁢

(

x

)

we find that

n

x

divides

n

. If we put

q

:=

|

Z

⁢

(

D

)

|

, we see
that

q

≥

2

since

{

0

,

1

}

⊂

Z

⁢

(

D

)

, and that

|

C

D

⁢

(

x

)

|

=

q

n

x

and

|

D

|

=

q

n

.

It suffices to show that

n

=

1

to prove that multiplication is
abelian, since then

|

Z

⁢

(

D

)

|

=

|

D

|

and so

Z

⁢

(

D

)

=

D

.

We now consider

D

*

:=

D

-

{

0

}

and apply the

conjugacy class formula

.

|

D

*

|

=

|

Z

(

D

*

)

|

+

∑

x

[

D

*

:

C

D

*

(

x

)

]

which gives

q

n

-

1

=

q

-

1

+

∑

x

q

n

-

1

q

n

x

-

1

.

By Zsigmondy’s

theorem

, there exists a prime

p

that divides

q

n

-

1

but doesn’t divide any of the

q

m

-

1

for

0

<

m

<

n

, except in 2
exceptional cases which will be dealt with separately. Such a prime

p

will divide

q

n

-

1

and each of the

q

n

-

1

q

n

x

-

1

. So
it will also divide

q

-

1

which can only happen if

n

=

1

.

We now deal with the 2 exceptional cases. In the first case

n

equals

2

, which would

D

is a vector space of dimension 2 over

Z

⁢

(

D

)

, with elements of the form

a

+

b

⁢

α

where

a

,

b

∈

Z

⁢

(

D

)

. Such elements clearly commute so

D

=

Z

⁢

(

D

)

which contradicts
our

assumption

that

n

=

2

. In the second case,

n

=

6

and

q

=

2

. The

class equation

reduces to

64

-

1

=

2

-

1

+

∑

x

2

6

-

1

2

n

x

-

1

where

n

x

divides 6. This gives

62

=

63

⁢

x

+

21

⁢

y

+

9

⁢

z

with

x

,

y

and

z

integers

, which is impossible since the

right hand side

is

divisible

by 3 and the left hand side isn’t.

Title

proof of Wedderburn’s theorem

Canonical name

ProofOfWedderburnsTheorem

Date of creation

2013-03-22 13:10:50

Last modified on

2013-03-22 13:10:50

Owner

lieven (1075)

Last modified by

lieven (1075)

Numerical id

8

Author

lieven (1075)

Entry type

Proof

Classification

msc 12E15

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