Boolean Constructions and Discriminator Varieties
Semisimple Varieties
A semisimple variety has all its subdirectly irreducibles simple, so Birkhoff's decomposition lands directly on simple pieces. It is the hypothesis discriminator varieties satisfy and most varieties do not.
- Define semisimplicity for a variety.
- Contrast with the module-theoretic notion of semisimplicity.
- Show every discriminator variety is semisimple.
- Determine whether the converse holds.
- Relate semisimplicity to residual smallness and residual finiteness.
- Identify semisimple varieties among standard examples.
01The definition
A variety is semisimple when every subdirectly irreducible member is simple.
Birkhoff's theorem says every algebra is a subdirect product of subdirectly irreducibles. In a semisimple variety those irreducibles are simple, so every algebra is a subdirect product of simple algebras — decomposition terminates on genuinely atomic pieces rather than on objects that still have internal congruence structure.
Without semisimplicity the subdirectly irreducibles can be arbitrarily complicated, each with its own congruence lattice above the monolith, and the decomposition tells you much less.
02Comparison with module semisimplicity
A semisimple variety in the universal-algebraic sense does not have every member a direct product of simple algebras — only a subdirect product. Boolean algebras are semisimple, and an infinite Boolean algebra is not a direct product of copies of 2. Reading across from module theory without checking will produce false statements.
03Discriminator varieties are semisimple
- input: discriminator variety V with discriminator term t
- let A ∈ V be subdirectly irreducible with monolith μ
- in the Boolean product representation, A has a single stalk
- (a subdirectly irreducible Boolean product must be concentrated at a point)
- the stalks of a Boolean product in a discriminator variety are simple
- therefore A is simple
- conclusion: every SI member of V is simple, so V is semisimple
So semisimplicity is one component of the discriminator variety package, and it is the component that makes the Boolean product representation land on simple stalks.
04The converse fails
Semisimplicity alone does not give a discriminator variety. The additional ingredients are arithmeticity and the congruence extension property.
| Condition | Semisimple | Discriminator |
|---|---|---|
| Every SI is simple | Yes | Yes |
| Arithmetical | Not required | Yes |
| Congruence-distributive | Not required | Yes |
| Congruence extension property | Not required | Yes |
| Boolean product of simples | Not guaranteed | Yes |
| Common discriminator term | Not required | Yes |
The gap is substantial. A variety can be semisimple while failing congruence distributivity, in which case none of the Boolean machinery applies. Semisimplicity is a necessary condition for the discriminator theory rather than a sufficient one.
05Residual smallness and finiteness
Two related conditions bound the size of the subdirectly irreducibles rather than their congruence structure.
Semisimplicity and residual smallness are independent. A variety can have all its subdirectly irreducibles simple and yet have simple algebras of unbounded size; and it can have small subdirectly irreducibles that are not simple. Both conditions together are what finitely generated discriminator varieties supply.
06Examples
| Variety | Semisimple? | Note |
|---|---|---|
| Boolean algebras | Yes | One SI, namely 2, and it is simple. |
| Discriminator varieties | Yes | By definition of the package. |
| Modules over a semisimple ring | Yes | Agrees with the classical notion. |
| Lattices | No | There are SI lattices that are not simple. |
| Heyting algebras | No | SI Heyting algebras have a top-adjacent structure and are generally not simple. |
| Groups | No | SI groups need not be simple. |
| Abelian groups | No | The Prüfer groups are SI and not simple. |
| Distributive lattices | Yes | The two-element lattice is the only SI. |
The distributive lattice case is instructive: like Boolean algebras it has exactly one subdirectly irreducible, which is simple, so the variety is semisimple and every member is a subdirect power of the two-element chain — the Birkhoff representation of distributive lattices as rings of sets.
Frequently asked
Does semisimple imply congruence-distributive?
No. The two conditions are independent. Semisimplicity constrains which algebras are subdirectly irreducible; distributivity constrains the shape of every congruence lattice. Neither implies the other, though discriminator varieties satisfy both.
Is a semisimple variety necessarily residually small?
No. There can be simple algebras of unbounded cardinality in a semisimple variety, in which case the subdirectly irreducibles are not bounded. Residual smallness is an extra hypothesis and is what the finite generation assumption supplies in the discriminator case.
Why are Heyting algebras not semisimple?
Because a subdirectly irreducible Heyting algebra is characterised by having a second largest element, and such an algebra generally has proper non-trivial congruences. Only the two-element Heyting algebra is simple. This is why Heyting algebras are arithmetical without forming a discriminator variety.
- S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, Millennium Edition (a corrected re-typesetting of Springer GTM 78, 1981).
- G. Grätzer, Universal Algebra, 2nd edition, Springer.
- R. McKenzie, G. McNulty and W. Taylor, Algebras, Lattices, Varieties, Volume I.
Original KEVOS® explanatory article. Written from the topic map of the cited works; no text is reproduced from them.