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By algebraic theory I mean one in the sense of Lawvere, i.e. a collection of finitary operations, including projections, together with a multi-composition satisfying the obvious axioms. (I believe universal algebraists call these abstract clones?)

A strict terminal object in a category is a terminal object $1$ such that every morphism with domain $1$ is an isomorphism. A well-known example of a category with strictly terminal objects is the category of (unital) rings: the only ring homomorphisms with domain $\{ 0 \}$ are isomorphisms.

It is a fact that any model of an algebraic theory whose underlying set is a singleton must be a terminal object in the corresponding category of models. My question is,

When are terminal objects in the category of models for an algebraic theory strict?

It is easy to generalise the proof that the algebraic theory of rings has the strict terminal object property: indeed, for any algebraic theory with two constants $0$ and $1$ and a binary operation for which $0$ is a (left) absorbing element and $1$ is a (left) unit, the trivial model is strictly terminal.

Another easy observation is that any algebraic theory that "interprets" an algebraic theory with the strict terminal object property must itself have the strict terminal object property. (Here, $T$ interprets $T'$ if there is a map from the operations of $T'$ to the operations of $T$ that preserves the multi-composition and projections.) So a natural follow-up question is,

Is there a "universal" algebraic theory with the strict terminal object property?

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I believe these would be the algebraic theories where there is a finite set of identities of closed terms that together imply all identities. (If there are no closed terms, just add a fresh constant symbol with no new axioms.) This is the case for rings, where the single identity $0 = 1$ implies that all ring identities hold.

It is easy to see that if there is such a finite set of identities of closed terms, then the terminal algebra $1$ must be strict. For the converse, let $\mathcal F$ be the collection of all finite sets of identities of closed terms. For each $\Sigma \in \mathcal F$, suppose that $A_\Sigma$ is a nontrivial algebra that satisfies $\Sigma$. Let $\mathfrak{U}$ be an ultrafilter on $F$ that contains all the sets $\{\Sigma \in \mathcal F : a=b \in \Sigma\}$, where $a = b$ denotes a fixed identity between closed terms $a$ and $b$. The ultraproduct $A^\ast = \prod_{\Sigma \in \mathcal F} A_\Sigma/\mathfrak{U}$ is nontrivial since each $A_\Sigma$ is nontrivial. However, it does admit a homomorphism from from the terminal algebra $1$ since all closed terms evaluate to the same thing in $A^\ast$ and hence the map sending the unique element of $1$ to that element of $A^\ast$ is a homomorphism.


Adding a new constant symbol works when the theory has no closed terms becase the terminal algebra is strict if and only if it is the only algebra with a trivial subalgebra. The new constant can be thought as a generator for a subalgebra and the criterion above boils down to the existence of finitely many identities of that generator that together imply that the whole algebra is trivial.

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I'm not sure I completely understand your construction. But I think there is an equivalent construction using compactness: adjoin to the theory the (first-order) axiom $\exists x . \exists y . x \ne y$ and all equations of closed terms; every finite subset is consistent, so this extension is also consistent. Thus the required non-trivial algebra exists. However, how do we deal with algebraic theories that have no closed terms at all? –  Zhen Lin Mar 3 at 12:31
    
@ZhenLin: My construction is a the special case of the ultrafilter proof of compactness for the theory you have in mind, so the two are the same. You can always add a free constant, with no axioms, if there are no closed terms; that doesn't affect anything. –  François G. Dorais Mar 3 at 12:36
    
@ZhenLin: I added a brief explanation for the case with no closed terms. Yet another way to handle that case is to use terms with one fixed variable symbol $x$. Everything works the same except that one must be careful to understand implication as meaning $\forall x(a = b \to c = d)$ and not $\forall x(a = b) \to \forall x(c = d)$. In fact, with this interpretation of implication, the "closed term" restriction can be removed entirely since the same works for any fixed finite number of variable symbols. –  François G. Dorais Mar 3 at 13:16

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