# Equational theories determined by “identities without variables”

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?

-
Another way to ask this question: when is the initial algebra of a variety existentially closed? – François G. Dorais Mar 8 '14 at 17:26
@FrançoisG.Dorais This sounds very interesting but I do not feel well with this notion, could you please explain some more? I mean I know the definition but it feels like stronger than what I said... – მამუკა ჯიბლაძე Mar 8 '14 at 19:00
No not even that, I am now perfectly confused. What happens, say, in the algebras over an algebraically closed field? Are they all subquotients of products of the field?? What about noncommutative algebras, is not the base field existentially closed there? – მამუკა ჯიბლაძე Mar 8 '14 at 19:08
The notion of existential closedness depends on the class you're working in. An algebraically closed field is existentially closed in the class of fields, but not in the class of algebras over that field (for example, an algebra over $K$ may have nilpotent elements, while $K$ does not). – Alex Kruckman Mar 11 '14 at 16:33
@AlexKruckman I see, thank you. I really feel uncertain about this notion... – მამუკა ჯიბლაძე Mar 11 '14 at 16:41

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.

-

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.

-
Sorry I feel this answer holds the key but don't quite understand why :D You established a correspondence between what and what? On one side you seemingly have all clones on any sets containing all constant maps from these sets to themselves. On the other side you seemingly have certain algebras in certain varieties. Do you mean that these algebras are in one-to-one correspondence with varieties in question? Specifically I am puzzled by your third line. I would somehow presume that the algebra is itself free on an empty set in the corresponding to be pinned down variety, no? – მამუკა ჯიბლაძე Sep 7 '15 at 5:52
@მამუკაჯიბლაძე My first appearance of the word "algebra" in the third line was supposed to be "variety", but I mistyped it. Does that clear everything up? – Will Sawin Sep 7 '15 at 13:45
I am on the verge of accepting it :D It took me some time to convince myself and then some more to convince myself that no further details need to be added. There remains just a minor question regarding examples - there might be further identities hidden, no? At least with the first example, as was pointed to me today at the seminar, one seemingly actually gets the subvariety of distributive bounded lattices... – მამუკა ჯიბლაძე Sep 9 '15 at 16:22
@მამუკაჯიბლაძე Yes, I was not sure I got all the identities. I guess you can check that you get all of them by checking that every algebra embeds into a proudct of copies of $\{0,1\}$. Wikipedia just told me that every bounded distributive lattice is a lattice of sets, so that's fine. Clearly every $\mathbb F_2$-vector space lives in a product of copise of $\mathbb F_2$. With join or with meet orders there's a similar embedding into sets (this time respecting only the join or only the meet). The last two are easy. – Will Sawin Sep 9 '15 at 18:34
Seems to be fine. Except some might not be embeddable into a product of copies of the initial algebra but rather be a subquotient. But all free ones should be so embeddable - in fact, the collection of all $n$-ary functions from your clone must be the free algebra on $n$ generators ($=$ projections). – მამუკა ჯიბლაძე Sep 10 '15 at 12:06