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I'd like to know more about varieties (in the sense of universal algebra) where every algebra is free. Another way to state the condition is that the comparison functor from the Kleisli category to the Elienberg-Moore category is an equivalence. For example, every object is free in

  • The category of sets

  • The category of pointed sets

  • The category of vector spaces (over a specified field), or more generally, the category of modules over a division ring

  • (Added 3/14/14) The category of affine spaces (vector spaces without a zero) and affine maps (linear maps + translations) over a division ring $R$. The algebraic structure is given by, for each $r \in R$, a ternary operation $f_r(x,y,z)$ meaning essentially $r(x-y)+z$, with appropriate relations to specify this. In the vector space case this example is mentioned in the paper John Baldwin links to below.

Is there a name for this property? Over at the n-Category Café, Zhen Lin suggested the term "absolutely free", but I gather this has a different meaning in universal algebra.

Has this property been studied in the literature? Are there other good examples? It seems like a very restrictive condition: is it restrictive enough to obtain some kind of structure theory for varieties with this property?

In the commutative algebra case: If all the modules over a ring $k$ are free, then is $k$ necessarily a division ring?

EDIT (2/19/14) The Masked Avenger mentions below that this property can be parsed in terms of categoricity in the sense of model theory. This reminds me that on the n-Category Café, Zhen Lin mentioned there should be an approach in terms of elimination of imaginaries. If anybody could flesh out the model-theoretic aspects I'd really appreciate it. Perhaps the topic has been well-covered model-theoretically?

I think the linear case has been clarified by multiple people. Benjamin Steinberg has some interesting results related to the classification aspect; any further observations would be great. I'm still looking for a name for this property[3/14/14: "pantofree" sounds joke-y to my ear, but maybe it is apt after all...], and still looking for further interesting examples. Maybe I'll also mention: one variation that might be interesting is to require only that finitely generated algebras be free.

Tim Campion
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    pantofree? panto means "every" (as in pantomime or pantograph) – YCor Feb 18 '14 at 21:25
  • I've spoken too soon. (Been working with finite types too much.) I have retracted the comment. Gerhard "Needs More Abstract Linear Algebra" Paseman, 2014.02.18 – Gerhard Paseman Feb 18 '14 at 21:34
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    To answer Tim's last question: note that a quotient ring of a commutative ring $A$ becomes an $A$-module. So the answer is yes for that case, since maximal two-sided ideals exist. – Todd Trimble Feb 18 '14 at 21:36
  • @ToddTrimble: That sounds pretty convincing -- I'll have to convince myself that a maximal quotient ring can't be a free module. – Tim Campion Feb 18 '14 at 23:34
  • This may be overkill, but to expand on my last comment: given any commutative ring quotient map $A \to k$, if $k$ were a free $A$-module, then there would exist a surjective $A$-module map $k \to A$. Now compose these to get a surjective $A$-module map $A \to A$. Show that any such is an isomorphism (if $a$ maps to $1$, show $a$ is invertible). Hence $A \to k$ was an iso to begin with. Since all ring quotients are isos, it must be that the only proper ideal of $A$ is $0$, so $A$ must be a field. – Todd Trimble Feb 19 '14 at 01:08
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    In model theory, it seems the theory of the variety is $\lambda$-categorical for those cardinals $\lambda$ strictly larger than the cardinality of the language used for the similarity type. You might check the model theory literature. – The Masked Avenger Feb 19 '14 at 04:40
  • In a variety each free algebra is projective, and the projectives are precisely the retracts of the frees. So if every object is free then every object is projective - which is something. – john Feb 19 '14 at 09:32
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    The question about modules over rings was previously asked here: http://mathoverflow.net/questions/515/rings-over-which-every-module-is-free – Manny Reyes Feb 19 '14 at 15:24
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    A trivial observation: "inconsistent" algebraic theories have only free algebra, all of which are even finitely generated. Excluding those cases, the trivial algebra can only be generated by zero or one element, which implies there is at most one constant (up to provable equality) in an algebraic theory whose algebras are free. (Recall that the free functor preserves monomorphisms.) – Zhen Lin Feb 20 '14 at 10:17
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    Saying that the Kleisli and Eilenberg-Moore categories are equivalent is a good way of formalising the question, but what appears to be missing from it is the monad. (Actually, that suggests that some abstract category theory could be brought to bear on the question.) – Paul Taylor Mar 02 '14 at 21:43
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    Mixing Greek and English morphemes is uncanny. If anything, it should be panteleutheric (ἐλεύθερος = free). – Emil Jeřábek Mar 18 '14 at 15:44

4 Answers4

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If there are no constant symbols in the language, $\mathcal L$, then the nontrivial varieties where every algebra is free are exactly the varieties term equivalent to sets or to affine spaces over a division ring. Here is why.

  1. If there are no constants in the language, $F_{\mathcal V}(\emptyset)$ is empty, so $F_{\mathcal V}(1)$ must be the 1-element algebra. This forces the variety to be idempotent.

  2. $\mathcal V$ is $\kappa$-categorical for $\kappa\geq |\mathcal L|$. This forces the first-order theory of infinite members of $\mathcal V$ to be complete, so $\mathcal V$ is a minimal variety.

  3. The minimal idempotent varieties were partially classified in my paper

Almost all minimal idempotent varieties are congruence modular, Algebra Universalis 44 (2000), 39-45.

Each one must be equivalent to the variety of sets, the variety of semilattices, a variety of affine modules over a simple ring, or must be congruence distributive.

  1. So what is left is to rule out the nonabelian ones and to show that a variety of affine modules over a simple ring where ever member is free is actually affine over a division ring. The division ring conclusion can be reached in various ways, e.g., by referring to the answer by Mariano Suárez-Alvarez. (You can shorten that argument a bit by noting that $\mathcal V$ has a simple member, and for a simple module to be free the ring must be a division ring.)

What is still left is to rule out the varieties equivalent to semilattices and the congruence distributive varieties. This can be done directly or we can cite papers of Baldwin + coauthors.

A variety equivalent to semilattices is locally finite, so by his paper with Lachlan (cited in Baldwin's answer), which applies to locally finite varieties whose infinite members are free, $\mathcal V$ is totally categorical. These varieties are known. A congruence distributive variety is congruence modular, so by his paper with McKenzie called "Counting models in universal Horn classes", Algebra Universalis 15 (1982), 359-384, the $\kappa$-categoricity and the congruence modularity of $\mathcal V$ jointly imply that $\mathcal V$ is abelian. (Their paper assumes a countable language, but this part of their argument doesn't require it.)

I don't know how to do the case where there are constants in the language. What can be shown is that there is at most one constant up to equivalence, that $F_{\mathcal V}(1)$ is abelian and simple, and that $\mathcal V = SPP_U(F_{\mathcal V}(1))$ is a minimal variety that is minimal as a quasivariety. The only examples I know are the varieties of pointed sets and the varieties of vector spaces over a division ring.

EDIT (8/16/15) I know more now. Steve Givant solved the main question posed here (Which varieties have the property that all members are free?) in his 1975 PhD thesis. The answer is: only those varieties term equivalent to sets, pointed sets, vector spaces over a division ring or affine spaces over a division ring. His methods do not seem to apply to the variation suggested by Tim: For which varieties are the finitely generated members free? However Emil Kiss, Agnes Szendrei and I just worked out the answer to that question (it is the same answer): sets, pointed sets, vectors spaces and affine spaces. I just submitted a short note on this to the arxiv.

Keith Kearnes
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    Thanks, this is really neat! A few questions: First, I'm confused by the first sentence of the second-to-last paragraph (A variety equivalent to semilattices...) -- doesn't it suffice here to observe that there are non-free semilattices? I also find your last paragraph tantalizing -- how does one conclude that $F_\mathcal{V}(1)$ is abelian and so forth? – Tim Campion Jul 07 '15 at 20:32
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    (1) Yes, it suffices to show that there are nonfree semilattices. (2) In the last paragraph I tried to copy what was done in the case where there are no constants, and some of it still goes through. The abelianness of $F_{\mathcal V}(1)$ follows from the Baldwin-McKenzie paper, for example. (More explicitly: if a variety has a nonabelian algebra it has a nonabelian subdirectly irreducible algebra, which B&M show is cancellable in Boolean powers. This is enough to guarantee that $\mathcal V$ has the maximum number of models in large cardinalities.) – Keith Kearnes Jul 08 '15 at 08:51
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    I felt like this answer merited a mention here at MO success stories: http://meta.mathoverflow.net/a/2385/2926 – Todd Trimble Aug 18 '15 at 08:42
  • No, finitely generated BA's are projective, but not necessarily free. A BA must have size $2^n$ for some $n$, and there is exactly one isomorphism type of size $2^n$ for any given $n$, but only those with $n$ equal to a power of $2$ are free. – Keith Kearnes Aug 20 '15 at 03:28
  • @KeithKearnes (After edit) This is fantastic! (and surprising to me -- I was thinking that restricting the condition to the finitely generated case might allow for interesting stuff to happen in the infinitely-generated algebras). I'm confused by one thing, though -- Zhen Lin Low points out here that every finitely generated boolean algebra is free. How do boolean algebras fit in? – Tim Campion Aug 20 '15 at 04:09
  • @TimCampion Sorry to interpose myself, but I don't see how Zhen Lin is right about that. The free Boolean on $n$ elements has $2^{2^n}$ elements, whereas a finitely generated Boolean has $2^k$ elements for some $k$ which we may choose to be not a power of $2$. – Todd Trimble Aug 20 '15 at 05:05
  • Okay, thanks. I should really stop to think a bit about boolean algebras (By the way, the comments above appear out of order because I deleted my comment and re-posted it with a slight edit, not realizing that Keith Kearnes had already replied to it. A word of warning to anyone trying similar fancy footwork!). – Tim Campion Aug 20 '15 at 05:59
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    @TimCampion The way I often think of free Boolean rings is the same way I think of polynomial algebras: just as a polynomial algebra is a free (additive) $k$-module on (the underlying set of) a free monoid, so a Boolean ring is a free $\mathbb{Z}/(2)$-module on a free commutative idempotent monoid. The free commutative idempotent monoid or meet-semilattice on $n$ elements has $2^n$ elements, and the free $\mathbb{Z}/(2)$-module on that would have $2^{2^n}$ elements. – Todd Trimble Aug 20 '15 at 17:43
  • I'm going to start counting the number of times I forget how finite boolean algebras work and need to have it explained to me again on the internet. – Tim Campion Feb 04 '21 at 06:30
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The suggestion that this was studied by model theorist is well-founded. See MR0351785 (50 #4273) 02G20 08A15 02H05 Baldwin, J. T.; Lachlan, A. H. On universal Horn classes categorical in some innte power. Algebra Universalis 3 (1973), 98{111.

The main result is that if the variety is locally finite, even under the weak assumption that the theory of the infinite models is complete, the variety is categorical in all infinite cardinalities. The article above has historical background. As far as I know, the problem remains open if `locally finite' is omitted.

John Baldwin
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If $R$ is a ring all of whose left modules are free, then every short exact sequence of modules splits and $R$ is left semisimple. It follows that $R$ is a finite direct product of minimal ideals.

If in this factorization there is more than one factor, then clearly no free module can be simple. But those minimal ideals are free simple modules by our hypothesis, and this is absurd. It follows that $R$ has has no proper non-zero left ideals, so every element has a left inverse and, as usual, this implies that every element has an inverse.

$R$ is therefore a division ring.

Mariano Suárez-Álvarez
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Edit.
My original answer was wrong. The varieties of left zero semigroups and right zero semigroups have the property and are the largest semigroup ones. They have identity basis $xy=x$ and $yx=x$ resp. The trivial variety is the only other.

If V is a variety of semigroups in which all semigroups are free then the relatively free monogenic semigroup on 1 generator must be a single idempotent because the trivial semigroup is not 0-generated. It follows V satisfies $x^2=x$. All varieties satisfying this identity are classified and left zero and right zero are the only nontrivial ones with all semigroups relatively free.

More generally if there are no constants in the signature the free algebra on one generator must be trivial. This then implies that each algebra in the variety is idempotent, meaning satisfies $f(x,\ldots,x)=x$ for each operation.

Benjamin Steinberg
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    And so each term operation is idempotent. In the same type, a 2-generated algebra will be either simple or (some version of) hopfian. – The Masked Avenger Feb 19 '14 at 04:32
  • On the one hand, it's encouraging to see that a complete classification is possible in this test case. On the other hand, it's too bad that we don't really get any new examples out of it: if I understand correctly, the variety of left-zero semigroups is categorically equivalent to the category of sets. – Tim Campion Feb 20 '14 at 03:35
  • Yes, in fact there is a unique left zero structure on any set and any map is a homomorphism so I think the underlying sent functor is an isomorphism and not just an equivalence. – Benjamin Steinberg Feb 20 '14 at 13:51
  • @TheMaskedAvenger: I'd like to be able to show this, but I'm having trouble constructing the argument. Could you explain the line of reasoning? – Tim Campion Mar 19 '14 at 16:06
  • @TimCampion, it's possible I overlooked something. However, the premise is quite restrictive, so something similarly strong should hold. Any (homomorphic) image of an fg (finitely-generated) algebra is fg. With no constants in the type, the congruence lattice of an algebra in your variety is special: any image of the algebra has a limited number of choices, so the $[\theta,1]$ subinterval of the congruence lattice is isomorphic to one of the congruence lattices of that choice. Either the fg algebra is simple, or hopfian, or we can find a larger congruence that "captures" much (continued) – The Masked Avenger Mar 19 '14 at 16:36
  • Of the original, by e.g. mapping from a 2-gen. to 7-gen. to three-gen back to 2-generated algebra, by climbing up the congruence lattice. If the type has constants, you can probably say something similar, but need more care. – The Masked Avenger Mar 19 '14 at 16:39
  • The image of an fg algebra is fg, sure -- but what if it is not freely finitely generated? Can this possibility be excluded? Also, "$A$ is simple" means there are no nontrivial quotients of $A$ and "$A$ is hopfian" means that any quotient $A \to A$ is an isomorphism -- right? I'm having trouble understanding what is special and limited about the congruence lattice -- does it satisfy some special finiteness condition or something? About all I can see that's special about the quotients is that every quotient splits / is representable as the image of an idempotent endomorphism. – Tim Campion Mar 19 '14 at 17:30
  • I think you will get that if the variety is locally finite then the 2-generated free algebra is simple. Any proper quotient would have to be 1-generated hence trivial. – Benjamin Steinberg Mar 19 '14 at 18:22
  • Okay -- in the locally finite case I can see this. But I don't see why it should hold in general. In fact, in the general case I can't rule out the possibility that all the finitely generated algebras are isomorphic. – Tim Campion Mar 20 '14 at 20:44
  • Probably that can happen. – Benjamin Steinberg Mar 20 '14 at 20:59
  • @Tim, even if it does happen, the fact that all the algebras are free is what limits the shape of the congruence lattice. Let C(\lambda) be the congruence lattice of the free algebra in this variety on \lambda many generators. In my version of set theory, k > l means C(l) appears as an interval (with top element 1) in C(k). The inclusion is proper when k > 2 unless all the lattices are trivial. C(k) appearing as an interval of C(l) can also happen, but if l is less than the (max of omega and the) cardinality of the type, so also must k be. The free algebra may not be hopfian ... (continued) – The Masked Avenger Mar 23 '14 at 20:54
  • But if it is not trivial or simple it will have a proper homomorphic image which is isomorphic to itself. – The Masked Avenger Mar 23 '14 at 20:55
  • I saw an example in one of Mark Sapir's writings (the link is somewhere on MO) from Tarski and Jonsson, I think. P(l(x),r(x))=x, l(P(x,y))=x and r(P(x,y))=y. This variety has only infinite nontrivial algebras; It might be possible to tweak this to a variety with all members being free. – The Masked Avenger Mar 23 '14 at 21:19
  • The free Jonsson-Tarski algebra in 1-generator is infinite and there is no constant term so the trivial algebra is not free. – Benjamin Steinberg Mar 24 '14 at 13:33
  • If we add a constant term to the Jonsson-Tarski signature, fixed by all operations, that makes the trivial algebra free. But I think it will be difficult to turn this into a variety where all algebras are free. A Jonsson-Tarski algebra is a bijection between $X$ and $X \times X$, and any sort of fixed-point property will be a non-trivial relation. For example, there should be a "free Jonsson-Tarski algebra on $n$ fixed points of all operations" for any $n$, but these algebras won't be free. – Tim Campion Mar 25 '14 at 02:51
  • I think I see the "almost-Hopfian" point now. If we take a quotient of the free algebra on 2 generators, the result is free. If it's nontrivial, then it's free on $n\geq 2$ generators; we can identify all but two of them to obtain a further quotient which is free on two generators. Thanks! – Tim Campion Mar 25 '14 at 02:53