# Varieties where every algebra is free

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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pantofree? panto means "every" (as in pantomime or pantograph) –  YCor Feb 18 '14 at 21:25
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
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 4:40
The question about modules over rings was previously asked here: mathoverflow.net/questions/515/… –  Manny Reyes Feb 19 '14 at 15:24
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

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.

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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.

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«there is more than one factor» kills me. –  Mariano Suárez-Alvarez Feb 19 '14 at 6:18

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.

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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 4:32
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 3:35
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