Is a "smooth" finite-dimensional algebra separable modulo its radical? Let $k$ be a field, and let us write the "unadorned" tensor $\otimes$ in place of $\otimes_k$. For a unital finite-dimensional $k$-algebra $A$, let $A^e = A \otimes A^{op}$ denote the enveloping algebra, so that $k$-central $(A,A)$-bimodules are the same as left $A^e$-modules. 
Recall that a finite-dimensional $k$-algebra $A$ is separable if it satisfies the following equivalent conditions:


*

*$A$ is projective as a left $A^e$-module;

*$A \otimes E$ is semisimple for every field extension $E/k$;

*$A^e$ is semisimple;

*$A \cong \prod_{i=1}^n \mathbb{M}_n(D_i)$ for division algebras $D_i$ whose centers are separable field extensions of $k$.


The first item above can be restated as "$A$ has projective dimension 0 as a left $A^e$-module." Thus the following property can be seen as a higher-dimensional generalization of separability.  The terminology below is borrowed from "homologically smooth" (dg-)algebras.

Definition: Say that a finite-dimensional $k$-algebra $A$ is smooth (of dimension $d$) if $A$ has finite projective dimension (equal to $d$) as a left $A^e$-module. 

This condition is known to be equivalent to $A^e$ having finite global dimension, and to imply that $A$ itself has finite global dimension.
Let $J(A)$ denote the Jacobson radical of $A$. I am curious about the relationship between smoothness of $A$ and of $A/J(A)$.

Question: If a finite-dimensional $k$-algebra $A$ is smooth, is the semisimple algebra $S = A/J(A)$ necessarily separable?

In case the answer is negative, I would greatly appreciate any further conditions that could be imposed on a smooth algebra $A$ as above to ensure that $S$ is separable.
What little I do know is that if a finite-dimensional algebra $A$ is smooth and semisimple, then $A = S$ is separable. (Indeed, both $A$ and $A^{op}$ are finite-dimensional semisimple, hence Frobenius, which implies that $A^e$ is Frobenius. It follows that every left $A^e$-module has projective dimension equal to $0$ or $\infty$. Since $A$ has finite $A^e$-projective dimension, it must be $A^e$-projective, so $A$ is separable.) In case the answer is negative, I would prefer a much better sufficient condition for $S$ to be separable, which allows for examples $A$ that have positive global dimension!
 A: Let $K$ be an algebraic closure of $k$.
The following lemma must surely be well-known, but I haven't found an explicit reference, so I'll include a proof at the end of this post.

Lemma. If $S$ is a finite dimensional semisimple $k$-algebra, then $S_K=S\otimes_kK$ is a finite product of matrix rings over commutative local $K$-algebras.

Let $A$ be as in the question and let $A_K=A\otimes_kK$. Applying $-\otimes_kK$ to a projective $A^e$-module resolution of $A$ gives a projective $A_K^e$-module resolution of $A_K$, so $A_K$ is also smooth and in particular has finite global dimension.
To prove that $S=A/J(A)$ is separable, it suffices to prove that $S_K=S\otimes_kK$ is semisimple.
By the lemma, $\operatorname{Ext}^1_{S_K}(U,V)=0$ for non-isomorphic simple $S_K$-modules $U$ and $V$.
If there is a non-split extension of $U$ by itself for some simple $S_K$-module $U$, then $\operatorname{Ext}^1_{A_K}(U,U)\neq0$. But since $A_K$ is a finite dimensional algebra of finite global dimension over an algebraically closed field, the "no loops conjecture" implies that this can't be the case. [Kiyoshi Igusa, MR 1086558 Notes on the no loops conjecture, J. Pure Appl. Algebra 69 (1990), no. 2, 161--176.]
Since there are no non-split extensions of simple $S_K$-modules, $S_K$ is semisimple.

Proof of the lemma. Let $K_s$ be the separable closure of $k$ in $K$.
Extending scalars to a separable field extension preserves semisimplicity, so $S_s=S\otimes_kK_s$ is semisimple, and therefore a finite product of matrix rings over division algebras finite dimensional over $K_s$. Since $K/K_s$ is purely inseparable, and hence $K/L$ is purely inseparable for any $L$ with $K_s\subseteq L\subseteq K$, all of these division algebras have separably closed centres. Since the Brauer group of a separably closed field is trivial, $S_s$ is a finite product of matrix rings over finite field extensions of $K_s$.
The tensor product of any field extension with a purely inseparable field extension is a commutative local algebra [see for example, Theorem 8.46(3) in Jacobson's Basic Algebra II]. So $S_K=S_s\otimes_{K_s}K$ is a finite product of matrix rings over commutative local $K$-algebras.
