reduced direct product


Let {AiiI} be a set of algebraic systems of the same type, indexed by I. Let A be the direct productMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of the Ai’s. For any a,bA, set

supp(a,b):={kIa(k)b(k)}.

Consider a Boolean ideal L of the Boolean algebraMathworldPlanetmath P(I) of I. Define a binary relationMathworldPlanetmath ΘL on A as follows:

(a,b)ΘL iff supp(a,b)L.
Lemma 1.

ΘL defined above is a congruence relationPlanetmathPlanetmath on A.

Proof.

Since L is an ideal L. Therefore, (a,a)ΘL, since {kIa(k)a(k)}=. Clearly, ΘL is symmetricPlanetmathPlanetmath. For transitivity, suppose (a,b,(b,c)ΘL. If a(k)c(k) for some kI, then either a(k)b(k) or b(k)c(k) (a contrapositive argument). So

supp(a,c)supp(a,b)supp(b,c).

Since L is an ideal, supp(a,c)L, so (a,c)ΘL, and ΘL is an equivalence relationMathworldPlanetmath on A.

Next, let ω be an n-ary operator on A and ajbj(modΘL), where j=1,,n. We want to show that ω(a1,,an)ω(b1,,bn)(modΘL). Let ωi be the associated n-ary operators on Ai. If ω(a1,,an)(k)ω(b1,,bn)(k), then ωk(a1(k),,an(k))ωk(b1(k),,bn(k)), which implies that aj(k)bj(k) for some j=1,,n. This implies that

supp(ω(a1,,an),ω(b1,,bn))j=1nsupp(aj,bj).

Since L is an ideal, and each supp(aj,bj)L, we have that supp(ω(a1,,an),ω(b1,,bn))L as well, this means that ω(a1,,an)ω(b1,,bn)(modΘL). ∎

Definition. Let A={AiiI}, L be a Boolean ideal of P(I) and ΘL be defined as above. The quotient algebra A/ΘL is called the L-reduced direct product of Ai. The L-reduced direct product of Ai is denoted by L{AiiI}. Given any element aA, its image in the reduced direct product L{AiiI} is given by [a]ΘL, or [a] for short.

Example. Let A=A1××An, and let L be the principal idealMathworldPlanetmathPlanetmathPlanetmathPlanetmath generated by 1. Then L={,{1}}. The congruencePlanetmathPlanetmathPlanetmath ΘL is given by (a1,,an)(b1,,bn)(modΘL) iff {iaibi}= or {1}. This implies that ai=bi for all i=2,,n. In other words, ΘL is isomorphicPlanetmathPlanetmathPlanetmath to the direct product of A2××An. Therefore, the L-reduced direct product of Ai is isomorphic to A1.

The example above can be generalized: if JI, then

{AiiI}P(J){AiiI-J}.

For aA={AiiI}, write a=(ai)iI. It is not hard to see that the map f:P(J){AiiI}{AiiI-J} given by f([a])=(ai)iI-J is the required isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.

Remark. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filter F. All there is to do is to replace supp(a,b) by its complementPlanetmathPlanetmath: supp(a,b)c:={kIa(k)=b(k)}. The congruence relation is now ΘF, where F={I-JJF} is the ideal complement of F. When F is prime, the F-reduced direct product is called a prime product, or an ultraproduct, since any prime filter is also called an ultrafilterMathworldPlanetmath. Ultraproducts can be more generally defined over arbitrary structuresMathworldPlanetmath.

References

Title reduced direct product
Canonical name ReducedDirectProduct
Date of creation 2013-03-22 17:10:11
Last modified on 2013-03-22 17:10:11
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 10
Author CWoo (3771)
Entry type Definition
Classification msc 08B25
Synonym ultraproduct
Defines prime product