standard duality on modules over algebras

standard duality on modules over algebras

Let

k

be a field and let

A

be an associative unital algebra. Throughout we will assume that all

A

-modules over

k

are unital. If

M

is a right

A

-module, then the space of all linear mappings

Hom

k

⁢

(

M

,

k

)

can be equipped with a left

A

-module

structure

as follows: for any

f

∈

Hom

k

⁢

(

M

,

k

)

and

a

∈

A

put

(

a

⁢

f

)

⁢

(

x

)

=

f

⁢

(

x

⁢

a

)

.

Note that action direction need to be reversed, because

(

a

⁢

b

⁢

f

)

⁢

(

x

)

=

(

b

⁢

f

)

⁢

(

x

⁢

a

)

=

f

⁢

(

x

⁢

a

⁢

b

)

.

Analogously

Hom

k

⁢

(

-

,

k

)

takes left

A

-modules to right

A

-modules. Also this action is

compatible

with functoriality of

Hom

k

⁢

(

-

,

k

)

, which means that it takes

A

-

homomorphisms

to

A

-homomorphisms. In particular we obtain a (contravariant)

functor

from

category

of left (right)

A

-modules to category of right (left)

A

-modules. Obviously

Hom

does not change the

dimension

of spaces, so we have well defined functors

D

:

mod

⁢

A

→

A

⁢

mod

D

:

A

⁢

mod

→

mod

⁢

A

which are

restrictions

of

Hom

(here

mod

means finite dimensional modules left/

right modules

) and are known in literature as ,,standard dualities”.

Proposition

.

Both

D

’s are quasi

inverse

dualities of categories.

Proof.

Let

M

be a finite dimensional

A

-module. We need to define a

natural isomorphism

between

M

and

D

⁢

D

⁢

(

M

)

. Indeed, define

τ

M

:

M

→

D

⁢

D

⁢

(

M

)

;

τ

M

⁢

(

m

)

⁢

(

α

)

=

α

⁢

(

m

)

.

We will show that each

τ

is an

isomorphism

.

1.

First we will show that

τ

is a

monomorphism

. Assume that

τ

M

⁢

(

m

)

=

0

for nonzero

m

∈

M

. This is if and only if

α

⁢

(

m

)

=

0

for every linear mapping

α

:

M

→

k

. But

m

is nonzero, so there is a basis of

M

(as

linear space

) which contains

m

. In particular there is a linear mapping

f

:

M

→

k

such that

f

⁢

(

m

)

=

1

.

Contradiction

. Thus

m

=

0

, which

completes

this part.

2.

τ

is an

epimorphism

. Indeed, let

F

:

D

⁢

(

M

)

→

k

be a linear mapping. We need to show, that there is

m

∈

M

such that

F

⁢

(

α

)

=

α

⁢

(

m

)

for any

α

∈

D

⁢

(

M

)

. Since

M

is finite dimensional, then let

{

e

1

,

…

,

e

n

}

be a

k

-basis of

M

. Of course

{

e

1

*

,

…

,

e

n

*

}

is a basis of

D

⁢

(

M

)

, where

e

i

*

is given by

e

i

*

⁢

(

e

j

)

=

1

if

i

=

j

and

e

i

*

⁢

(

e

j

)

=

0

otherwise. Define

λ

i

=

F

⁢

(

e

i

*

)

and put

m

=

∑

i

=

1

n

λ

i

⋅

e

i

.

We leave it as a simple exercise, that

τ

⁢

(

m

)

=

F

.

What remains is to prove, that

τ

is natural. Consider an

A

-homomorphism

f

:

X

→

Y

. We need to show that the following

diagram

commutes:

\xymatrix

⁢

X

⁢

\ar

⁢

[

r

]

f

⁢

\ar

⁢

[

d

]

τ

X

⁢

&

⁢

Y

⁢

\ar

⁢

[

d

]

τ

Y

⁢

D

⁢

D

⁢

(

X

)

⁢

\ar

⁢

[

r

]

D

⁢

D

⁢

(

f

)

⁢

&

⁢

D

⁢

D

⁢

(

Y

)

Indeed, if

x

∈

X

, then let

F

=

τ

X

⁢

(

x

)

. We have that

D

⁢

D

⁢

(

f

)

⁢

(

F

)

=

F

∘

D

⁢

(

f

)

and evaluating this at

α

∈

D

⁢

(

M

)

we have

F

⁢

(

D

⁢

(

f

)

⁢

(

α

)

)

=

F

⁢

(

α

∘

f

)

=

(

α

∘

f

)

⁢

(

x

)

=

α

⁢

(

f

⁢

(

x

)

)

=

τ

Y

⁢

(

f

⁢

(

x

)

)

⁢

(

α

)

.

In particular we obtain that

D

⁢

D

⁢

(

f

)

⁢

(

τ

X

⁢

(

x

)

)

=

τ

Y

⁢

(

f

⁢

(

x

)

)

which means that

D

⁢

D

⁢

(

f

)

∘

τ

X

=

τ

Y

∘

f

which completes the proof.

□

Title

standard duality on modules over algebras

Canonical

name

StandardDualityOnModulesOverAlgebras

Date of creation

2013-12-11 15:25:39

Last modified on

2013-12-11 15:25:39

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Theorem

Classification

msc 16S99

Classification

msc 20C99

Classification

msc 13B99

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