If $R$ is a commutative ring (with $1$), then an $R$-algebra $A$ is said to be *separable* if $A$ is projective as an $A$-$A$-bimodule. (The notion of an "$A$-$A$-bimodule" includes the requirement that $R$ acts the same way from the left and from the right. The notion of "projective $A$-$A$-bimodule" is defined in the same way as the notions of "projective $A$-left module" and "projective $A$-right module". If you are unhappy with this definition, you can rewrite any $A$-$A$-bimodule as an $A\otimes_R A^{\mathrm{op}}$-left module, and then use the notion of a projective left module.)

There is a criterion stating that an $R$-algebra $A$ is separable if and only if there is an element $e\in A\otimes_R A$ such that the multiplication map $A\otimes_R A\to A$ sends $e$ to $1\in A$, and such that $ae=ea$ for all $a\in A$. Equivalently, an $R$-algebra $A$ is separable if and only if the $A$-$A$-bimodule epimorphism $A\otimes_R A\to A$ given by multiplication of the two tensorands has a section in the category of $A$-$A$-bimodules. (This is both in Crawley-Boevey, chapter 4. I have difficulties finding other literature which does the notion of separability in full generality. For some reason, most books consider it enough to talk about separable $k$-algebras with $k$ a field.)

Now my question is, is there a transitivity theorem like this:

If $B$ is a separable commutative $R$-algebra, and $A$ is a separable $B$-algebra, then $A$ is a separable $R$-algebra as well?

Maybe some conditions like projectivity (of $B$ as $A$-module and $A$ as $R$-module) must be added; that would be ok for me.

If something like this holds, then the proof that separability of a field extension as algebra is just Galois-theoretical separability could be simplified (most importantly, the ugly Primitive Element Theorem would not be needed anymore).