Boolean Constructions and Discriminator Varieties
Skew-Free Algebras and Independence
Algebras whose congruences on a product decompose into products of congruences on the factors, and the independence conditions that force this.
Learning objectives
- Define skew congruences and skew-free algebras
- State when products are skew-free
- Connect skew-freeness to the structure of congruence lattices
Skew congruences
A congruence on a direct product A1 × … × An that is not of the form θ1 × … × θn for congruences θi on the factors.
A product is skew-free if it has no skew congruences — every congruence on the product is a product of congruences on the factors.
The natural guess is that congruences on a product should decompose. Skew congruences are the counterexamples, and they are common.
A skew congruence
Take Z/2 × Z/2 as a group. The diagonal subgroup {(0,0), (1,1)} is normal, so it determines a congruence. That congruence is not a product of congruences on the factors — the only candidates would be Δ × Δ, Δ × ∇, ∇ × Δ and ∇ × ∇, none of which is the diagonal congruence.
When products are skew-free
In a congruence-distributive variety, any finite direct product of algebras is skew-free provided the factors have no common non-trivial homomorphic images. In particular a product of pairwise non-isomorphic simple algebras is skew-free.
The argument uses distributivity of the congruence lattice to split a congruence along the factor congruences. In a merely modular setting the split need not occur, which is exactly why the group example above admits a skew congruence — groups are modular but not distributive.
| Variety | Products skew-free? |
|---|---|
| Boolean algebras | Yes |
| Distributive lattices | Yes |
| Any discriminator variety | Yes |
| Groups | No — the diagonal example |
| Rings | No |
| Modules | No |
Independence
Varieties V1 and V2 of the same type are independent if there is a term t with V1 satisfying t(x, y) ≈ x and V2 satisfying t(x, y) ≈ y.
If V1 and V2 are independent, then every algebra in the join variety V1 ∨ V2 is uniquely a direct product of an algebra in V1 and one in V2.
The witnessing term acts as a projection selector: it picks the first coordinate in one variety and the second in the other, which is exactly what is needed to separate the factors.
Why this matters
Skew-freeness and independence are the technical conditions under which direct decomposition behaves as one would naively expect. When they hold, the structure of a product is fully determined by the factors, and the subvariety lattice decomposes correspondingly.
A Boolean product representation is useful precisely because it approximates skew-freeness: the patchwork condition ensures that congruences of the whole are controlled by congruences of the stalks. Discriminator varieties, being congruence-distributive and semisimple, satisfy the strongest form of this.
Frequently asked questions
Is skew-freeness preserved under subalgebras?
Not in general. A subalgebra of a skew-free product can have congruences that do not extend, so skew-freeness is a property of the specific product rather than an inherited one.
Are independent varieties common?
Not especially. Independence requires a term behaving in opposite ways in the two varieties, which is a strong demand. When it holds, the payoff is a complete decomposition, which is why the condition is worth isolating.
Source. S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, The Millennium Edition — a corrected re-typesetting of Springer-Verlag Graduate Texts in Mathematics 78 (1981). Section IV.11, book pages 204-207.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates and reorganises mathematical results; it is not a reproduction of the source text.