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 $A$ issmooth(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!