direct product of partial algebras

Let 𝑨 and 𝑩 be two partial algebraic systems of type Ο„. The direct productMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of 𝑨 and 𝑩, written 𝑨×𝑩, is a partial algebra of type Ο„, defined as follows:

  • β€’

    the underlying set of 𝑨×𝑩 is AΓ—B,

  • β€’

    for each n-ary function symbol fβˆˆΟ„, the operationMathworldPlanetmath f𝑨×𝑩 is given by:

    for (a1,b1),…,(an,bn)∈AΓ—B, f𝑨×𝑩⁒((a1,b1)⁒…,(an,bn)) is defined iff both f𝑨⁒(a1,…,an) and f𝑩⁒(b1,…,bn) are, and when this is the case,


It is easy to see that the type of 𝑨×𝑩 is indeed Ο„: pick a1,…,an∈A and b1,…,bn∈B such that f𝑨⁒(a1,…,an) and f𝑩(b1,…,bn)) are defined, then f𝑨×𝑩⁒((a1,b1),…,(an,bn)) is defined, so that f𝑨×𝑩 is non-empty, where all operations are defined componentwise, and the two constants are (0,0) and (1,1).

For example, suppose k1 and k2 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 k1Γ—k2, while no longer a field, is still an algebraMathworldPlanetmathPlanetmath of the same type.

Let 𝑨,𝑩 be partial algebras of type Ο„. Can we embed 𝑨 into 𝑨×𝑩 so that 𝑨 is some type of a subalgebraMathworldPlanetmathPlanetmath of 𝑨×𝑩?

For example, if we fix an element b∈B, then the injection ib:𝑨→𝑨×𝑩, given by ib⁒(a)=(a,b) is in general not a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath only unless b is an idempotentMathworldPlanetmathPlanetmath with respect to every operation f𝑩 on B (that is, f𝑩(b,…,b)=b). In addition, b would have to be the constant for every constant symbol in Ο„. Following from the example above, if we pick any r∈k2, then r would have to be 0, since, (s1+s2,2⁒r)=ir⁒(s1)+ir⁒(s2)=ir⁒(s1+s2)=(s1+s2,r), so that 2⁒r=r, or r=0. But, on the other hand, ir⁒(s-1)=(s,r)-1=(s-1,r-1), forcingMathworldPlanetmath r to be invertiblePlanetmathPlanetmath, a contradictionMathworldPlanetmathPlanetmath!

Now, suppose we have a homomorphism Οƒ: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)).


Suppose f𝑨⁒(a1,…,an) is defined. Since Οƒ is a homomorphism, f𝑩⁒(σ⁒(a1),…,σ⁒(an)) is defined, which means f𝑨×𝑩⁒((a1,σ⁒(a1)),…,(an,σ⁒(an)))=f𝑨×𝑩⁒(ϕ⁒(a1),…,ϕ⁒(an)) is defined. Furthermore, we have that

f𝑨×𝑩⁒((a1,σ⁒(a1)),…,(an,σ⁒(an))) = (f𝑨⁒(a1,…,an),f𝑩⁒(σ⁒(a1),…,σ⁒(an)))
= (f𝑨⁒(a1,…,an),σ⁒(f𝑨⁒(a1,…,an)))
= ϕ⁒(f𝑨⁒(a1,…,an)),

showing that Ο• is a homomorphism. In addition, if f𝑨×𝑩⁒(ϕ⁒(a1),…,ϕ⁒(an)) is defined, then it is clear that f𝑨⁒(a1,…,an) is defined, so that Ο• is a strong homomorphism. So ϕ⁒(𝑨) is a subalgebra of 𝑨×𝑩. Clearly, Ο• 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 {π‘¨π’Šβˆ£i∈I} be a set of partial algebras of type Ο„, indexed by set I. The direct product of these algebras is a partial algebra 𝑨 of type Ο„, defined as follows:

  • β€’

    the underlying set of 𝑨 is A:=∏{Ai∣i∈I},

  • β€’

    for each n-ary function symbol fβˆˆΟ„, the operation f𝑨 is given by: for a∈A, f𝑨⁒(a) is defined iff fπ‘¨π’Šβ’(a⁒(i)) is defined for each i∈I, and when this is the case,


Again, it is easy to verify that 𝑨 is indeed a Ο„-algebra: for each symbol fβˆˆΟ„, the domain of definition dom⁑(fπ‘¨π’Š) is non-empty for each i∈I, and therefore the domain of definition dom⁑(f𝑨), being ∏{dom⁑(fπ‘¨π’Š)∣i∈I}, is non-empty as well, by the axiom of choiceMathworldPlanetmath.


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