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.

pantomimeorpantograph) – YCor Feb 18 '14 at 21:25panteleutheric(ἐλεύθερος = free). – Emil Jeřábek Mar 18 '14 at 15:44