Classification of separable algebras over a commutative ring A separable algebra over a field can be classified as a finite product of matrix algebras over division algebras whose centers are finite dimensional separable field extensions of the field. (See http://ncatlab.org/nlab/show/separable+algebra.)
Suppose we instead have a separable algebra over a commutative ring R. What is the classification result in this case? I am particularly interested in the case that the algebra is finitely generated and projective as an R module.
 A: I can give a very limited answer: the case when $R$ is semisimple ring.
Mimicking the proof of the classification theorem over fields, let $S$ be an $R$-algebra and $N$ be a (left) $S$-submodule of $S$. As $R$ is semisimple, there exists an$R$-submodule $N' \leq S$ such that
$S = N \oplus N'$
Let be $\pi : S \to S$ be the $R$-linear projector onto $N$ in this decomposition.
I believe you can upgrade it to an $S$-linear one by taking
$\pi' = \mu(-, (\mu \circ (\mathrm{id}_S \otimes_R \pi))   p)$
where $\mu : S \otimes_R S \to S$ is your algebra multiplication, and $p \in S^{op} \otimes_R S$ is your separability idempotent. The fact that $\pi'$ is an $S$-linear projection onto $N$ can be verified even with string diagrams.
This shows that $S$ is a semisimple $S$-module.
You still have a version of Schur's lemma which you can use to get that $S$ is a product of matrices in division algebras over $R$.
Caveats:


*

*All of this needs to be carefully checked

*There isn't an abundance of commutative, semisimple rings: they're just products of fields

*You have to work harder to get uniqueness

