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It might be a naive question, as I am not a specialist in this field. This is a follow-up to this question.

I want to study varieties of objects generalizing ordered monoids, in particular using an explicit operator $\omega$, subject to the axiom $(A)$ $xx=x\implies x^\omega=x$, i.e. the restriction of $\omega$ to idempotents has to be the identity function.

Does it make sense to talk about a free algebra in such a context? The problem is that this free algebra would be the same as the one where we don't specify axiom $(A)$, and replace it by axioms $(A')$ like $x^\omega=x^\omega x^\omega$ and $(x^\omega)^\omega=x^\omega$.

If two algebraic structures have exactly the same free object, it seems like this object does not really represent the constraints given by the axioms, and therefore the axioms are not of a shape that allows to define a free object. Is this reasoning valid?

For now, the approach taken is to use axioms $(A')$, and to later use a profinite equation $x^\omega=x^\pi$, where $\pi$ is the profinite idempotent power, in order to define the pseudovariety of objects we are really interested in. But I'm wondering if such a detour is necessary.

EDIT

More detailed question: If we were to use axiom (A), what is needed to be proved in order to show that there is a free object? It seems that closure under arbitrary product is important, why is it the case? Also, if we just want the universality property for finite objects, then is it enough to prove closure under finite products? Also, isn't enough to take as free object the set of terms quotiented by the axioms, and directly show that it has the universal property?

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  • $\begingroup$ You do need infinite products unless the variety generated by your objects is locally finite. For example, the class of finite Heyting algebras is closed under finite products, but does not have free objects over any nonempty set of generators, as these would need to be infinite Heyting algebras. $\endgroup$
    – Emil Jeřábek
    Commented Jan 25, 2014 at 22:34
  • $\begingroup$ ok, is it because using "finite" as an axiom seems even worse than using implications? And is there a precise theorem stating "closed under product+ something => exists free object" ? Birkhoff seems close but I'm not sure it is the one... $\endgroup$
    – Denis
    Commented Jan 25, 2014 at 22:39
  • $\begingroup$ As M. Shahryari mentioned, closure under products and subalgebras implies the existence of free objects over any set of generators. That’s a theorem, I don’t know what “precise” means (if anything). $\endgroup$
    – Emil Jeřábek
    Commented Jan 25, 2014 at 22:55
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    $\begingroup$ It seems to me there is one pretty trivial answer to the title question, so it must not answer what you want. An algebraic theory $T$ can be construed as a monad on $Set$, and given the category of free $T$-algebras together with the forgetful functor $U: Free_T \to Set$, with left adjoint $F: Set \to Free_T$, the monad $T$ can be recovered as the composite $U F$ (with the monad data recoverable from the adjunction data). On another question: free $T$-algebras exist for any monad $T$ on $Set$, and certainly varieties defined by operations and equations of bounded rank give monads. $\endgroup$
    – Todd Trimble
    Commented Jan 25, 2014 at 23:57
  • $\begingroup$ @ToddTrimble It is probably what I'm looking for, but I'm not very familiar with category theory. If I understand your comment, you are saying that having a free object is given almost for free according to category theory, as long as you are a monad. I tried to understand what a monad is on wikipedia, but it seems circular, because it looks like it is the structures having free objects... Also you mention varieties, but here it's not a variety, because axioms are not restricted to identities. $\endgroup$
    – Denis
    Commented Jan 26, 2014 at 0:28

2 Answers 2

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I think you seems to be interested in quasivarieties or quasipseudovarieties.

Let me stick to the non-pseudo world. For the pseudo-setting you need to always use profinite analogs.

While a variety is determined by its free object on a countable generating set (assuming all operations have finite arity) quasivarieties are highly non determined by free objects. For instance the cancellative laws are quasiidentities and so cancellative monoids form a quasivariety. But the free cancellative monoid is the same as the free monoid so the free object in the much larger quasivariety of all monoids is already free in the much smaller quasivariety if cancellative monoids.

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  • $\begingroup$ thanks this is useful. So in this case how do you show that the free object indeed has the universal property? $\endgroup$
    – Denis
    Commented Jan 26, 2014 at 15:43
  • $\begingroup$ The free monoid obviously has the desired property to be the free cancellative monoid. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 26, 2014 at 16:55
  • $\begingroup$ The free left regular band is free in uncountably many quasivarieties of bands. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 26, 2014 at 16:55
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    $\begingroup$ @PeterLanguilla can you explain your terminology $\endgroup$
    – Benjamin Steinberg
    Commented Jun 13, 2021 at 22:38
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    $\begingroup$ Quasivarieties are also closed under ultraproducts. In any event cancelative monoid and all monoids are ISP classes with the same free object. A variety is determined by its free object on a countable generating set because an identity has only finitely many variables. From this it is easy to see that any identity satisfied by the free object on $\omega$ of satisfied by all algebras in the variety $\endgroup$
    – Benjamin Steinberg
    Commented Jun 14, 2021 at 0:53
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Two non-isomorphic algebras $A$ and $B$ may generate the same variety and so the free objects of the variety generated by an algebra $A$ do not characterize $A$ uniquely. For example, for any group $A$, we have $Var(A)=Var(A\times A)$, so free objects are the same however $A$ and $A\times A$ are not isomorphic. In general the classes of algebras like the one you are dealing with are quasi-varieties and so they may not contain free objects (for example, note that there is no free field, since the class of fields is not a variety, it is just a quasi-variety). But it seems that the special class you introduced is a pre-variety too (it is closed under subalgebra and product) so it contains free objects (you should check it out if the class is really closed under product or not). For more details see Burris, Sankapanavar: A course in universal algebra.

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  • $\begingroup$ "Algebraic structure" in the title means something like "groups" or "monoids" in general, and not particular instances of them. $\endgroup$
    – Denis
    Commented Jan 25, 2014 at 21:02
  • $\begingroup$ I think we are only interested in closure under finite products. $\endgroup$
    – Denis
    Commented Jan 25, 2014 at 21:04
  • $\begingroup$ This is not sufficient, if you want to have free objects inside your class, you must control that is it closed under arbitrary product or not. I think that it is closed. $\endgroup$
    – Sh.M1972
    Commented Jan 25, 2014 at 21:06
  • $\begingroup$ But still, it wouldn't be a problem that the free (A)-algebra with k generators is the same as the free (A')-algebra with k generators? $\endgroup$
    – Denis
    Commented Jan 25, 2014 at 21:24
  • $\begingroup$ @M._Shahryari what is the link between free objects and infinite products? $\endgroup$
    – Denis
    Commented Jan 25, 2014 at 21:44

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