Idempotent complete Kleisli category Are there some Kleisli categories in which all idempotents split?
The answer is yes: let $C$ be an idempotent complete category, the Kleisli category of the identity monad is $C$ itself. Are there some other, more interesting examples?

Already in the case of monads on $\textbf{Set}$ this is a remarkable property.
Recall that monads on $\textbf{Set}$ can be identified with (possibly infinitary) algebraic theories.
Under this identification, the Kleisli category is identified with the category of free algebras.
Therefore "idempotents in the Kleisli category split" is equivalent to "every retract of a free algebra is free".
Put another way, since the retracts of free algebras are precisely the regular-projective algebras, the property in question is equivalent to "every regular-projective algebra is free".
This tells us immediately that there are many examples and non-examples occurring "in nature".
Example. Every vector space over a field $k$ has a basis, i.e. every vector space is free as an algebra.
So the Kleisli category of the free $k$-vector space monad is idempotent-complete.
(In fact, it is even complete.)
Example. Every projective abelian group is free.
So the Kleisli category of the free abelian group monad is idempotent-complete.
Non-example. Let $A$ be the ring of $2 \times 2$ matrices over a field $k$.
Then $k^2$ considered as a left $A$-module is projective but not free.
So the Kleisli category of the free left $A$-module monad is not idempotent-complete.
(Incidentally, the category of left $A$-modules is abstractly equivalent to the category of $k$-vector spaces, so every left $A$-module is projective.)
