free objects in the category of commutative algebras

free objects in the category of commutative algebras

Let

R

be a commutative ring and let

𝒜

⁢

ℒ

⁢

𝒢

c

⁢

(

R

)

be the

category

of all

commutative algebras

over

R

and algebra homomorphisms. This category together with the

forgetful functor

is a

construct

(i.e. it is a concrete category over the

category of sets

𝒮

⁢

ℰ

⁢

𝒯

). Therefore we can talk about free objects in

𝒜

⁢

ℒ

⁢

𝒢

c

⁢

(

R

)

(see this entry (

http://planetmath.org/FreeObjectsInConcreteCategories2

) for definitions).

Theorem

.

For any set

𝕏

the

polynomial algebra

R

⁢

[

𝕏

]

(see parent object) is a free object in

𝒜

⁢

ℒ

⁢

𝒢

c

⁢

(

R

)

with

𝕏

being a basis. This means that for any commutative algebra

A

and any

function

f

:

𝕏

→

A

there exists a unique algebra homomorphism

F

:

R

⁢

[

𝕏

]

→

A

such that

F

⁢

(

x

)

=

f

⁢

(

x

)

for any

x

∈

𝕏

.

Proof.

Assume that

f

:

𝕏

→

A

is a function. If

W

∈

R

⁢

[

𝕏

]

, then there are finite subsets

A

1

,

…

,

A

n

⊆

𝕏

(not necessarily

disjoint

) and

natural numbers

n

⁢

(

x

,

i

)

,

i

=

1

,

…

,

n

such that

W

can be uniquely expressed as

W

=

∑

i

=

1

n

(

λ

i

⋅

∏

x

∈

A

i

x

n

⁢

(

x

,

i

)

)

with

λ

i

∈

R

. Define

F

⁢

(

W

)

by putting

F

⁢

(

W

)

=

∑

i

=

1

n

(

λ

i

⋅

∏

x

∈

A

i

f

⁢

(

x

)

n

⁢

(

x

,

i

)

)

.

Of course

F

is well defined and obviously

F

⁢

(

x

)

=

f

⁢

(

x

)

. We leave as a simple exercise that

F

is an algebra homomorphism.
The uniqueness of

F

again follows from the explicit form of

W

. It is easily seen that

F

⁢

(

W

)

depends only on

F

⁢

(

x

)

for

x

∈

𝕏

. This

completes

the proof.

□

Corollary 1.

If

𝕏

is a set and

𝕐

⊆

𝕏

, then the

inclusion

i

:

𝕐

→

𝕏

induces an

algebra

monomorphism

I

:

R

⁢

[

𝕐

]

→

R

⁢

[

𝕏

]

.

In particular we can treat

R

⁢

[

𝕐

]

as a

subalgebra

of

R

⁢

[

𝕏

]

.

Proof.

We have a well-defined function

i

:

𝕐

→

R

⁢

[

𝕏

]

,

i

⁢

(

y

)

=

y

. By the theorem we have an

extension

I

:

R

⁢

[

𝕐

]

→

R

⁢

[

𝕏

]

such that

I

⁢

(

y

)

=

y

. It remains to show, that

I

is ,,1-1”. Indeed, assume that

I

⁢

(

W

)

=

0

for some

polynomial

W

∈

R

⁢

[

𝕐

]

. But if we recall the expression of

W

as in proof of the theorem and remember that

I

is an algebra homomorphism, then it is easy to see that

I

⁢

(

y

)

=

y

implies that

I

⁢

(

W

)

=

W

.

In particular

W

=

0

, which completes the proof.

□

Corollary 2.

If

A

is an

R

-algebra, then there exists a set

𝕏

such that

A

≃

R

⁢

[

𝕏

]

/

I

for some ideal

I

.

Proof.

Let

𝕏

=

A

as a set. Define

f

:

𝕏

→

A

by

f

⁢

(

x

)

=

x

. By the theorem we have an algebra homomorphism

F

:

R

⁢

[

𝕏

]

→

A

such that

F

⁢

(

x

)

=

x

for

x

∈

𝕏

. In particular

F

is ,,onto” and thus by the

First Isomorphism Theorem

for algebras we have

A

≃

R

⁢

[

𝕏

]

/

Ker

⁢

F

which completes the proof.

□

Title

free objects in the category of commutative algebras

Canonical name

FreeObjectsInTheCategoryOfCommutativeAlgebras

Date of creation

2013-03-22 19:18:13

Last modified on

2013-03-22 19:18:13

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Theorem

Classification

msc 13P05

Classification

msc 11C08

Classification

msc 12E05

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