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](http://www.amsta.leeds.ac.uk/~pmtwc/cohom.pdf), 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).



Is there any transitivity for separable algebras?
You do not even need that the intermediate algebra be commutative:



> 
> Let $R$ be commutative and let $A\subseteq C\subseteq B$ be a chain of $R$-algebras. If $B$ is a separable extension of $C$ and $C$ of $A$, then $B$ is a separable extension of $A$.
> 
> 
> 


This is one of the exercises in Pierce's *Associative Algebras*, §10.8. 




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**Later:** Here is an answer to darij's question in the comments below:


If two $R$-algebras $A$ and $B$ are Morita equivalent, so that their categories of left modules ${}\_A\mathrm{Mod}$ and ${}\_B\mathrm{Mod}$ are equivalent (as $R$-linear categories) then one can see that their categories of bimodules ${}\_A\mathrm{Mod}{}\_B$ and ${}\_B\mathrm{Mod}{}\_B$ are also equivalent (as $R$-linear categories, again) by an equivalence which maps the $A$-bimodule $A$ to the $B$-bimodule $B$. 


Now $A$ being a separable $R$-algebra means that the $A$-bimodule $A$ is a projective object in ${}\_A\mathrm{Mod}{}\_A$. Since projectivity of an object is preserved by equivalences of categories, we conclude that 



> 
> an $R$-algebra which is Morita equivalent to a separable $R$-algebra is itself separable.
> 
> 
> 


Now $A$ and $M\_n(A)$ are Morita equivalent $R$-algebras for all $n\geq1$, so an affirmative answer to the question follows.


One can even be bolder: if $A$ and $B$ are now only *derived* equivalent (but now I think we need to assume they are both flat over $R$...) then it also follows that they are simultaneously separable, by a similar reasoning.


