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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?

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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

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