Is there any transitivity for separable algebras? 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).
 A: Yes, separability should be transitive. Calling the element $e$ a separation element. If $\sum_i x_i\otimes y_i$ is a separation element for $B\to A$ and $\sum_j z_j\otimes w_j$ is a separation element for $R\to B$ then $\sum_{i,j}x_iz_j\otimes_R y_iw_j$ is a separation element for the composite arrow $R\to A$.
A: 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. 

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.
