Equational theories determined by "identities without variables"

Question

How to characterize equational theories $T$ which have the following property: for any two terms $t(x_1,...,x_n)$ and $t'(x_1,...,x_n)$ in the signature of $T$, if for any closed terms (i. e. terms without variables) $c_1$, ..., $c_n$ the identity $t(c_1,...,c_n)=t'(c_1,...,c_n)$ follows from (the identities of) $T$ then so does $t(x_1,...,x_n)=t'(x_1,...,x_n)$.

In algebraic terms this means to characterize those varieties of algebras which are generated by their initial algebra (the free algebra on the empty set).

The only widely known example of this that I was able to come up with is the theory/variety of Boolean algebras. But in fact any algebra $A$ in any signature generates its own variety of this kind, by adding to the signature a bunch of constants in the well known way and then generating the subvariety by $A$ itself.

So I am interested in any "intrinsic" (say, category-theoretic) characterization of theories/varieties with the above property, as well as in any other familiar examples of such.

May one hope to actually classify such things up to, say, categorical equivalence?

Much later

After so much time I hardly made any progress on this, but here is one feature which might be useful.

I believe I have a (simple) proof that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

Answer 1

I don't know the answer to this question, but I have a correction and a remark.

The question asks "When is a variety $\mathcal V$ generated by its free algebra over the empty set?" It has been asserted that this question is equivalent to the question of when ${\mathbf F}_{\mathcal V}(\emptyset)$ is existentially closed in $\mathcal V$.

The correction. The assertion is not true. There is an implication, but not an equivalence.

Let ${\mathbf F}_0:={\mathbf F}_{\mathcal V}(\emptyset)$, and assume that $\mathbf F_0$ is existentially closed (e.c.) in $\mathcal V$. $\mathbf F_0$ is embeddable is $\mathbf F_m:=\mathbf F_{\mathcal V}(m)$ for every $m$. If the q.f. formula $s(\bar{x})\neq t(\bar{x})$ is satisfiable in $\mathbf F_m$, then it is satisfiable in $\mathbf F_0$ by the e.c. property. This shows that $\mathbf F_0$ fails every identity that fails in $\mathcal V$, while the fact that $\mathbf F_0\in{\mathcal V}$ shows that it satisfies every identity that holds in $\mathcal V$. Altogether this shows that $\mathbf F_0$ e.c. implies $\mathcal V$ is generated by $\mathbf F_0$.

But the converse is false. The variety of commutative rings is generated by ${\mathbf F}_0 = \mathbb Z$, but this ring is not e.c. in the variety of commutative rings. Moreover, there are many nontrivial varieties generated by their initial algebras, $\mathbf F_0$, where these initial algebras happen to be finite. (E.g., the variety of Boolean algebras, or bounded distributive lattices, or the variety generated by the ring of integers modulo $n$, or any variety generated by the constant expansion of a finite algebra.) But a nontrivial finite algebra cannot be e.c.

The remark. The original poster asks if there is a category-theoretic characterization of these varieties. Well, there is one, since you can express Birkhoff's HSP Theorem category-theoretically. To express $\mathcal V = \mathbf{HSP}(\mathbf F_0)$ you just need to say that every object in $\mathcal V$ is the image of an extremal epimorphism from some object that has a monomorphism into some power of the initial algebra. But, I doubt that there is a nontrivial characterization of these varieties. As the original poster noted, any variety generated by the constant expansion of an algebra has the desired property, so this class of varieties is as varied as the class of constant expansions of algebras.

Answer 2

Aren't these the same as clones containing the constant functions ?

From a clone, you just take the equational theory consisting of all the functions in it, with equations all the relations satisfied by those functions.

From a theory/variety, just take all the operations restricted to the free algebra on the empty set in that varietey.

There are many nontrivial clones on a two-element set, but if I read the diagram right there are only finitely many containing the constant symbols. Among these are:

The clone of monotone functions gives you the variety of bounded lattices.

The clone of affine functions gives you the variety of $\mathbb F_2$-vector spaces with a marked point.

The clone of conjunctive or disjunctive functions gives you the variety of bounded partial orders with joins or with meets.

The clone of unary functions gives you the variety of sets with an involution and two marked elements that are switched by the involution.

The clone of all Boolean operations gives you the variety of Boolean algebras.

The clone of constant functions gives you the variety of sets with two marked elements.

Answer 3

This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one, $\mathbf{E}$, is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set. (There is a set containing at least one isomorphic copy of each subdirectly irreducible algebra in $\mathscr{V}$.) We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ cannot be represented by a set. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.