Suppose $F:C\to D$ is a conservative functor admitting a left adjoint $F^L:D\to C$. We obtain a monad $$ T=FF^L:D\to D. $$

Under the Barr--Beck hypotheses, we have $$ C\simeq \operatorname{Alg}_T(D). $$

On the other hand, we can construct the Kleisli category $K=\operatorname{Kl}(T)$, whose objects are those of $D$, with $$ \operatorname{Hom}_K(X,Y)=\operatorname{Hom}_D(X,TY). $$

The Kleisli category embeds fully faithfully into $\operatorname{Alg}_T(D)$ as the category of free $T$-algebras.

A natural question is:

If $D$ is already perfect, i.e. $D\simeq\operatorname{Perf}(D)$, do we necessarily have $$ \boxed{\operatorname{Perf}(K)\simeq\operatorname{Alg}_T(D)\,?} $$

The answer is no, even in a very elementary example.

Let $k$ be a field, and consider the algebra $$ A=k[\varepsilon]/(\varepsilon^2). $$

Take $$ D=\operatorname{Perf}(k), $$ the category of bounded complexes of finite-dimensional $k$-vector spaces.

In particular, $D$ is already stable and idempotent-complete: $$ D\simeq\operatorname{Perf}(D). $$

Define the monad $$ T(V)=A\otimes_k V. $$

Its unit and multiplication are induced by the unit and multiplication of $A$.

The Eilenberg--Moore category is $$ \boxed{ \operatorname{Alg}_T(D) \simeq D^b(\operatorname{mod}_{fd}A). } $$

Indeed, a $T$-algebra is an $A$-module whose underlying $k$-complex is perfect, or equivalently, an $A$-complex with bounded finite-dimensional cohomology.

On the other hand, the Kleisli category consists of free $A$-modules of the form $A\otimes_k V$, for $V\in\operatorname{Perf}(k)$.

Taking its perfect envelope gives $$ \boxed{ \operatorname{Perf}(\operatorname{Kl}(T)) \simeq\operatorname{Perf}(A) =K^b(\operatorname{proj}A). } $$

Thus the question becomes whether $$ \operatorname{Perf}(A) \overset{?}{=} D^b(\operatorname{mod}_{fd}A). $$

Consider the simple $A$-module $$ M=k=A/(\varepsilon), $$ on which $\varepsilon$ acts by zero.

Clearly $M$ is a finite-dimensional $k$-vector space, so it is an object of $\operatorname{Alg}_T(D)$.

However, $M$ is not a perfect $A$-module.

To see this, consider its free resolution: $$ \cdots \xrightarrow{\varepsilon} A \xrightarrow{\varepsilon} A \xrightarrow{\varepsilon} A \xrightarrow{\varepsilon} A \longrightarrow k \longrightarrow 0. $$

This complex is exact because $$ \ker(\varepsilon:A\to A) =(\varepsilon) =\operatorname{im}(\varepsilon:A\to A). $$

Thus $k$ has an infinite periodic free resolution.

In fact, applying $\operatorname{Hom}_A(-,k)$ gives $$ 0\longrightarrow k \xrightarrow{0}k \xrightarrow{0}k \xrightarrow{0}k \xrightarrow{0}\cdots, $$ because $\varepsilon$ acts trivially on $k$.

Consequently, $$ \boxed{ \operatorname{Ext}^n_A(k,k)\cong k \qquad\text{for every }n\geq 0. } $$

If $k$ were perfect over $A$, it would admit a bounded complex of finitely generated projective $A$-modules as a resolution. This would imply $$ \operatorname{Ext}^n_A(k,k)=0 \qquad\text{for }n\gg0, $$ a contradiction.

Therefore $$ k\notin\operatorname{Perf}(A). $$

We have exhibited an object $$ k\in\operatorname{Alg}_T(D) $$ which does not belong to the thick subcategory generated by free $T$-algebras.

Hence $$ \boxed{ \operatorname{Perf}(\operatorname{Kl}(T)) \subsetneq \operatorname{Alg}_T(D). } $$

This happens despite the fact that $D=\operatorname{Perf}(k)$ is already perfect, and the forgetful functor $$ F:D^b(\operatorname{mod}_{fd}A)\to\operatorname{Perf}(k) $$ is conservative and monadic.

The point: Being perfect after forgetting the $T$-algebra structure does not imply being perfect as a $T$-algebra.

The simple module $k$ is perfect as a $k$-complex, but has infinite projective dimension as an $A$-module.

A useful sufficient condition is that $A$ has finite global dimension.

Indeed, if $$ \operatorname{gldim}(A)<\infty, $$ then every finite-dimensional $A$-module admits a finite projective resolution, so $$ D^b(\operatorname{mod}_{fd}A) \simeq\operatorname{Perf}(A). $$

Thus, for a finite-dimensional $k$-algebra $A$, $$ \boxed{ \operatorname{Perf}(\operatorname{Kl}(A\otimes_k-)) \simeq \operatorname{Alg}_{A\otimes_k-}(\operatorname{Perf}(k)) } $$ whenever $A$ has finite global dimension.

The dual numbers provide a counterexample precisely because $$ \operatorname{gldim}(k[\varepsilon]/(\varepsilon^2))=\infty. $$

More generally, the correct condition is that every $T$-algebra can be obtained from free $T$-algebras using finitely many cones, shifts, and retracts.