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.



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> **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. 
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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 $A/J(A)$.



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> **Question:** If a finite-dimensional $k$-algebra $A$ is smooth, is the semisimple algebra $S = A/J(A)$ necessarily separable?
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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](https://math.stackexchange.com/questions/919135/tensor-product-of-algebras-which-is-frobenius) that $A^e$ is Frobenius. [It follows](https://math.stackexchange.com/questions/218721/global-dimension-of-quasi-frobenius-ring) 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!



Is a "smooth" finite-dimensional algebra separable modulo its radical?
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.



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> **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.
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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**](http://dx.doi.org/10.1016/0022-4049(90)90040-O), *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.




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*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.


