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Is there an algebra $A$ (for example a group) such that $Th(A)$ is logically equivalent to $id(A)$? In other words, is there an algebra $A$ such that $$ Mod(Th(A))=Var(A)? $$ Clearly finite algebras do not have this property. It seems that such an algebra should be relatively free.

This question is related to my previous two questions

  1. Relatively free algebras in a variety generated by a single algebra

  2. relatively free groups in $Var(S_3)$

Edition: Only trivial algebra has this property by the comment of Gerhard Paseman. So I ask again the question by $\pm id(A)$ instead of $id(A)$. Is there any algebra A (especially a group) such that $Th(A)$ is logically equivalent to $\pm id(A)$? Here $\pm id(A)$ means the set of identities and negated identities.

P.S. By negated identity I mean a sentence of the form $$ \forall x_1\ldots \forall x_n: p(x_1, \ldots, x_n)\neq q(x_1, \ldots, x_n), $$ where $p$ and $q$ are terms. Is there any negated identity in a non-trivial algebra? Clearly there are no negated identities in groups but if we add constants to the language of groups there will be many negated identities.

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  • $\begingroup$ Does $\text{id}(A)$ include only equations, or are negated equations also allowed? $\endgroup$
    – Joel David Hamkins
    Commented Jan 9, 2014 at 21:51
  • $\begingroup$ The trivial (one-element) algebra is an example. I doubt there are any others without an appropriately restricted definition of $Th(\textbf{A})$. Gerhard "Ask Me About System Design" Paseman, 2014.01.09 $\endgroup$
    – Gerhard Paseman
    Commented Jan 9, 2014 at 21:55
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    $\begingroup$ Isn't it true that the one-element algebra satisfies everything in id(A) for any A? In this case, it will not satisfy Th(A) unless A also has only one element. Thus, the only example is the trivial algebra. $\endgroup$
    – Joel David Hamkins
    Commented Jan 9, 2014 at 21:57
  • $\begingroup$ Trivially (wink) yes. Gerhard "If Th(A) Is As Expected" Paseman, 2014.01.09 $\endgroup$
    – Gerhard Paseman
    Commented Jan 9, 2014 at 22:00
  • $\begingroup$ Gerhard, why not post this as an answer? $\endgroup$
    – Joel David Hamkins
    Commented Jan 9, 2014 at 22:01

3 Answers 3

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I imagine that definitions of $Mod, Th, Var$ and so on have not changed since I saw them decades ago. The trivial one element algebra in any finite type (and likely any infinite type) is an easy example which satisfies $Mod(Th(\textbf{A})) \approx Var(\textbf{A})$. Since it is expected that $Th(\textbf{A})$ is strong enough to indicate whether $\textbf{A}$ has more than one element, this is the only example to be expected (Thank you Joel).

Gerhard "Ask Me About Trivial Algebra" Paseman, 2014.01.09

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For an example with a negated identity, let $A$ be a vector space over an infinite field in the usual signature ($+$ and scalar multiplications) together with an additional constant $1\ne0$.

EDIT: Since it was apparently not obvious (judging from the comment), this is meant to be an example of an algebra whose full first-order theory is equivalent to its set of valid identities and negated identities. (This follows here from the fact that the theory of infinite vector spaces over a field $F$ is categorical in every cardinality $\kappa>|F|$.)

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  • $\begingroup$ Here is another example: in the algebra $(\mathbb{R}, 0, 1, +, -, \times)$, we have the negated identity $x^2\neq -1$. $\endgroup$
    – Sh.M1972
    Commented Jan 10, 2014 at 13:39
  • $\begingroup$ That won’t work (and more generally, no ring will do). The set of identities and negated identities valid in the real field is satisfied in every formally real integral domain (and more), for example in $\mathbb Z$, which has a quite different first-order theory. $\endgroup$
    – Emil Jeřábek
    Commented Jan 10, 2014 at 14:35
  • $\begingroup$ Yes I know, I just gave an example of negated identity. $\endgroup$
    – Sh.M1972
    Commented Jan 10, 2014 at 15:32
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Here is another example. Let $A$ be the algebra consisting of infinitely many distinct constants $c_n$. The theory of this model is logically equivalent to the assertions $c_n\neq c_m$ for distinct $n,m$, and so the models of $\text{Th}(A)$ are the same as the models of $\pm\text{id}(A)$.

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  • $\begingroup$ How do you express “there are no other objects” without disjunction? As far as I can see, if $A=\langle\{0,1\},0,1\rangle$, then $\pm\mathrm{id}(A)$ holds in every $B\supseteq A$, even though they are not elementarily equivalent. $\endgroup$
    – Emil Jeřábek
    Commented Jan 10, 2014 at 15:31
  • $\begingroup$ In fact, the validity of identities and negated identities is preserved by nonempty Cartesian products, so no finite algebra with at least $2$ elements has the property. $\endgroup$
    – Emil Jeřábek
    Commented Jan 10, 2014 at 15:34
  • $\begingroup$ You are right. I was thinking of disjunctions of equations, but this is not allowed. I have edited. $\endgroup$
    – Joel David Hamkins
    Commented Jan 10, 2014 at 15:37
  • $\begingroup$ @Joel David Hamkins: in the case of groups (in any language without constants except 1), there are no negated identity. $\endgroup$
    – Sh.M1972
    Commented Jan 10, 2014 at 15:38
  • $\begingroup$ I had had in mind a language with a constant for the generator(s), but that example doesn't work anyway for the reasons that Emil mentioned. $\endgroup$
    – Joel David Hamkins
    Commented Jan 10, 2014 at 15:39

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