direct product of partial algebras

direct product of partial algebras

Let

𝑨

and

𝑩

be two

partial algebraic systems

of type

τ

. The

direct product

of

𝑨

and

𝑩

, written

𝑨

×

𝑩

, is a

partial algebra

of type

τ

, defined as follows:

•

the underlying set of

𝑨

×

𝑩

is

A

×

B

,

•

for each

n

-ary

function symbol

f

∈

τ

, the

operation

f

𝑨

×

𝑩

is given by:

for

(

a

1

,

b

1

)

,

…

,

(

a

n

,

b

n

)

∈

A

×

B

,

f

𝑨

×

𝑩

⁢

(

(

a

1

,

b

1

)

⁢

…

,

(

a

n

,

b

n

)

)

is defined iff both

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

and

f

𝑩

⁢

(

b

1

,

…

,

b

n

)

are, and when this is the case,

f

𝑨

×

𝑩

⁢

(

(

a

1

,

b

1

)

,

…

,

(

a

n

,

b

n

)

)

:=

(

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

,

f

𝑩

⁢

(

b

1

,

…

,

b

n

)

)

.

It is easy to see that the type of

𝑨

×

𝑩

is indeed

τ

: pick

a

1

,

…

,

a

n

∈

A

and

b

1

,

…

,

b

n

∈

B

such that

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

and

f

𝑩

(

b

1

,

…

,

b

n

)

)

are defined, then

f

𝑨

×

𝑩

⁢

(

(

a

1

,

b

1

)

,

…

,

(

a

n

,

b

n

)

)

is defined, so that

f

𝑨

×

𝑩

is non-empty, where all operations are defined componentwise, and the two

constants

are

(

0

,

0

)

and

(

1

,

1

)

.

For example, suppose

k

1

and

k

2

are fields. They are both partial algebras of type

⟨

2

,

2

,

1

,

1

,

0

,

0

⟩

, where the two

2

’s are the arity of

addition

and multiplication, the two

1

’s are the arity of

additive

and

multiplicative inverses

, and the two

0

’s are the constants

0

and

1

. Then

k

1

×

k

2

, while no longer a field, is still an

algebra

of the same type.

Let

𝑨

,

𝑩

be partial algebras of type

τ

. Can we embed

𝑨

into

𝑨

×

𝑩

so that

𝑨

is some type of a

subalgebra

of

𝑨

×

𝑩

?

For example, if we fix an

element

b

∈

B

, then the

injection

i

b

:

𝑨

→

𝑨

×

𝑩

, given by

i

b

⁢

(

a

)

=

(

a

,

b

)

is in general not a

homomorphism

only unless

b

is an

idempotent

with respect to every operation

f

𝑩

on

B

(that is,

f

𝑩

(

b

,

…

,

b

)

=

b

)

. In addition,

b

would have to be

the

constant for every constant symbol in

τ

. Following from the example above, if we pick any

r

∈

k

2

, then

r

would have to be

0

, since,

(

s

1

+

s

2

,

2

⁢

r

)

=

i

r

⁢

(

s

1

)

+

i

r

⁢

(

s

2

)

=

i

r

⁢

(

s

1

+

s

2

)

=

(

s

1

+

s

2

,

r

)

, so that

2

⁢

r

=

r

, or

r

=

0

. But, on the other hand,

i

r

⁢

(

s

-

1

)

=

(

s

,

r

)

-

1

=

(

s

-

1

,

r

-

1

)

,

forcing

r

to be

invertible

, a

contradiction

!

Now, suppose we have a homomorphism

σ

:

𝑨

→

𝑩

, then we may embed

𝑨

into

𝑨

×

𝑩

, so that

𝑨

is a subalgebra of

𝑨

×

𝑩

. The embedding is given by

ϕ

⁢

(

a

)

=

(

a

,

σ

⁢

(

a

)

)

.

Proof.

Suppose

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

is defined. Since

σ

is a homomorphism,

f

𝑩

⁢

(

σ

⁢

(

a

1

)

,

…

,

σ

⁢

(

a

n

)

)

is defined, which means

f

𝑨

×

𝑩

⁢

(

(

a

1

,

σ

⁢

(

a

1

)

)

,

…

,

(

a

n

,

σ

⁢

(

a

n

)

)

)

=

f

𝑨

×

𝑩

⁢

(

ϕ

⁢

(

a

1

)

,

…

,

ϕ

⁢

(

a

n

)

)

is defined. Furthermore, we have that

f

𝑨

×

𝑩

⁢

(

(

a

1

,

σ

⁢

(

a

1

)

)

,

…

,

(

a

n

,

σ

⁢

(

a

n

)

)

)

=

(

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

,

f

𝑩

⁢

(

σ

⁢

(

a

1

)

,

…

,

σ

⁢

(

a

n

)

)

)

=

(

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

,

σ

⁢

(

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

)

)

=

ϕ

⁢

(

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

)

,

showing that

ϕ

is a homomorphism. In addition, if

f

𝑨

×

𝑩

⁢

(

ϕ

⁢

(

a

1

)

,

…

,

ϕ

⁢

(

a

n

)

)

is defined, then it is clear that

f

𝑨

⁢

(

a

1

,

…

,

a

n

)

is defined, so that

ϕ

is a strong homomorphism. So

ϕ

⁢

(

𝑨

)

is a subalgebra of

𝑨

×

𝑩

. Clearly,

ϕ

is one-to-one, and therefore an embedding, so that

𝑨

is

isomorphic

to

ϕ

⁢

(

𝑨

)

, and we may view

𝑨

as a subalgebra of

𝑨

×

𝑩

.
∎

Remark

. Moving to the general case, let

{

𝑨

𝒊

∣

i

∈

I

}

be a set of partial algebras of type

τ

,

indexed by

set

I

. The

direct product

of these algebras is a partial algebra

𝑨

of type

τ

, defined as follows:

•

the underlying set of

𝑨

is

A

:=

∏

{

A

i

∣

i

∈

I

}

,

•

for each

n

-ary function symbol

f

∈

τ

, the operation

f

𝑨

is given by: for

a

∈

A

,

f

𝑨

⁢

(

a

)

is defined iff

f

𝑨

𝒊

⁢

(

a

⁢

(

i

)

)

is defined for each

i

∈

I

, and when this is the case,

f

𝑨

⁢

(

a

)

⁢

(

i

)

:=

f

𝑨

𝒊

⁢

(

a

⁢

(

i

)

)

.

Again, it is easy to verify that

𝑨

is indeed a

τ

-algebra: for each symbol

f

∈

τ

, the

domain of definition

dom

⁡

(

f

𝑨

𝒊

)

is non-empty for each

i

∈

I

, and therefore the domain of definition

dom

⁡

(

f

𝑨

)

, being

∏

{

dom

⁡

(

f

𝑨

𝒊

)

∣

i

∈

I

}

, is non-empty as well, by the

axiom of choice

.

References

1

G. Grätzer:

Universal Algebra

, 2nd Edition, Springer, New York (1978).

Title

direct product of partial algebras

Canonical name

DirectProductOfPartialAlgebras

Date of creation

2013-03-22 18:43:40

Last modified on

2013-03-22 18:43:40

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Definition

Classification

msc 08A55

Classification

msc 08A62

Classification

msc 03E99

Defines

direct product

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