ring of endomorphisms

ring of endomorphisms

Let

R

be a ring and let

M

be a right

R

-module.

An

endomorphism

of

M

is a

R

-

module homomorphism

from

M

to itself.
We shall write endomorphisms on the left,
so that

f

:

M

→

M

maps

x

↦

f

⁢

(

x

)

.
If

f

,

g

:

M

→

M

are two endomorphisms, we can add them:

f

+

g

:

x

↦

f

⁢

(

x

)

+

g

⁢

(

x

)

and multiply them

f

⁢

g

:

x

↦

f

⁢

(

g

⁢

(

x

)

)

With these

operations

, the set of endomorphisms of

M

becomes a ring, which we call
the

ring of endomorphisms

of

M

, written

End

R

⁡

(

M

)

.

Instead of writing endomorphisms as

functions

, it is often convenient
to write them multiplicatively: we simply write the application of the
endomorphism

f

as

x

↦

f

⁢

x

. Then the fact that each

f

is an

R

-module homomorphism can be expressed as:

f

⁢

(

x

⁢

r

)

=

(

f

⁢

x

)

⁢

r

for all

x

∈

M

and

r

∈

R

and

f

∈

End

R

⁡

(

M

)

.
With this notation, it is clear that

M

becomes an

End

R

⁡

(

M

)

-

R

-

bimodule

.

Now, let

N

be a left

R

-module. We can

construct

the ring

End

R

⁡

(

N

)

in the same way.
There is a complication, however, if we still think of endomorphism as functions written
on the left. In order to make

M

into a bimodule, we need to define an action of

End

R

⁡

(

N

)

on the right of

N

: say

x

⋅

f

=

f

⁢

(

x

)

But then we have a problem with the

multiplication

:

x

⋅

f

⁢

g

=

f

⁢

g

⁢

(

x

)

=

f

⁢

(

g

⁢

(

x

)

)

but

(

x

⋅

f

)

⋅

g

=

f

⁢

(

x

)

⋅

g

=

g

⁢

(

f

⁢

(

x

)

)

!

In order to make this work, we need to reverse the order of composition
when we define multiplication in the ring

End

R

⁡

(

N

)

when it acts on the right.

There are essentially two different ways to go from here. One is to define the multiplication
in

End

R

⁡

(

N

)

the other way, which is most natural if we write the endomorphisms as functions
on the right. This is the approach taken in many older books.

The other is to leave the
multiplication in

End

R

⁡

(

N

)

the way it is, but to use the opposite ring to define the bimodule.
This is the approach that is generally taken in more recent works. Using this approach,
we conclude that

N

is a

R

-

End

R

(

N

)

op

-bimodule. We will adopt this convention
for the lemma below.

Considering

R

as a right and a

left module

over itself,
we can construct the two

endomorphism rings

End

R

⁡

(

R

R

)

and

End

R

⁡

(

R

R

)

.

Lemma

.

Let

R

be a ring with an

identity element

.
Then

R

≅

End

R

⁡

(

R

R

)

and

R

≅

End

R

(

R

R

)

𝑜𝑝

.

Proof.

Define

ρ

r

∈

End

R

⁡

(

R

R

)

by

x

↦

x

⁢

r

.

A calculation shows that

ρ

r

⁢

s

=

ρ

s

⁢

ρ

r

(functions written on the left)
from which it is easily seen that the map

θ

:

r

↦

ρ

r

is a

ring
homomorphism

from

R

to

End

R

(

R

R

)

op

.

We must show that this is an

isomorphism

.

If

ρ

r

=

0

, then

r

=

1

⁢

r

=

ρ

r

⁢

(

1

)

=

0

. So

θ

is

injective

.

Let

f

be an arbitrary

element of

End

R

⁡

(

R

R

)

, and let

r

=

f

⁢

(

1

)

.
Then for any

x

∈

R

,

f

⁢

(

x

)

=

f

⁢

(

x

⁢

1

)

=

x

⁢

f

⁢

(

1

)

=

x

⁢

r

=

ρ

r

⁢

(

x

)

, so

f

=

ρ

r

=

θ

⁢

(

r

)

.

The proof of the other isomorphism is

similar

.
∎

Title

ring of endomorphisms

Canonical name

RingOfEndomorphisms

Date of creation

2013-03-22 14:04:26

Last modified on

2013-03-22 14:04:26

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

8

Author

mclase (549)

Entry type

Definition

Classification

msc 16S20

Classification

msc 16W20

Synonym

endomorphism ring

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