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.

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    $\begingroup$ pantofree? panto means "every" (as in pantomime or pantograph) $\endgroup$
    – YCor
    Feb 18, 2014 at 21:25
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    $\begingroup$ 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. $\endgroup$
    – Todd Trimble
    Feb 18, 2014 at 21:36
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    $\begingroup$ 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. $\endgroup$ Feb 19, 2014 at 4:40
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    $\begingroup$ The question about modules over rings was previously asked here: mathoverflow.net/questions/515/… $\endgroup$ Feb 19, 2014 at 15:24
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    $\begingroup$ Mixing Greek and English morphemes is uncanny. If anything, it should be panteleutheric (ἐλεύθερος = free). $\endgroup$ Mar 18, 2014 at 15:44

4 Answers 4


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.

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    $\begingroup$ 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? $\endgroup$
    – Tim Campion
    Jul 7, 2015 at 20:32
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    $\begingroup$ (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.) $\endgroup$ Jul 8, 2015 at 8:51
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    $\begingroup$ I felt like this answer merited a mention here at MO success stories: meta.mathoverflow.net/a/2385/2926 $\endgroup$
    – Todd Trimble
    Aug 18, 2015 at 8:42
  • $\begingroup$ 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. $\endgroup$ Aug 20, 2015 at 3:28
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    $\begingroup$ @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. $\endgroup$
    – Todd Trimble
    Aug 20, 2015 at 17:43

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 in nte 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.


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.

  • $\begingroup$ «there is more than one factor» kills me. $\endgroup$ Feb 19, 2014 at 6:18

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.

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    $\begingroup$ And so each term operation is idempotent. In the same type, a 2-generated algebra will be either simple or (some version of) hopfian. $\endgroup$ Feb 19, 2014 at 4:32
  • $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Feb 20, 2014 at 3:35
  • $\begingroup$ 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. $\endgroup$ Feb 20, 2014 at 13:51
  • $\begingroup$ @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? $\endgroup$
    – Tim Campion
    Mar 19, 2014 at 16:06
  • $\begingroup$ @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) $\endgroup$ Mar 19, 2014 at 16:36

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