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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).

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Can you take a look at the second part of the proof of proposition 2.3 here… – Gjergji Zaimi Jan 13 '11 at 11:40
Thankss for the link! Will look into it. Note that the paper is open access:… – darij grinberg Jan 13 '11 at 13:55
Something tells me that "as $\Lambda\otimes_C \Lambda^0$-modules" should be "as $\Lambda\otimes_R \Lambda^0$-modules", but otherwise the proof seems OK. Thanks again! – darij grinberg Jan 13 '11 at 19:12
Gjergji, one more thing: your first proposed solution WAS right! You can post it again, or if you don't, I will (tomorrow). – darij grinberg Jan 13 '11 at 22:26
up vote 3 down vote accepted

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.

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OK, now I need to know what "separable" means over a non-commutative algebra. Guess it's in Pierce. Thanks for the reference! – darij grinberg Jan 15 '11 at 17:49
If $A\subseteq B$ is an inclusion of $R$-algebras, $B$ is a separable extension of $A$ if the map $B\otimes_AB\to B$ given by multiplication has a section as a map of $B$-bimodules. – Mariano Suárez-Alvarez Jan 15 '11 at 18:05
Thanks a lot. Mariano, do you happen to know other results of this kind? Like, if $\mathrm M_n\left(A\right)$ is separable over $R$ (commutative) for some $n$, then so is $A$ ? – darij grinberg Jan 15 '11 at 20:47
Thanks! Somehow I didn't realize that Morita equivalence yields something for bimodules and not just for one-sided modules. – darij grinberg Jan 17 '11 at 0:42

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 $R\to A$.

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Why does your last tensor (let's call it $t$) satisfy $bt=tb$ for all $b\in B$? Note that the $\otimes$ sign in your first tensor mean $\otimes_A$, which means less information than $\otimes_R$ as in your last tensor. – darij grinberg Jan 13 '11 at 10:43
Oh! I forgot to update my comment to this one. In contrast to what I said before, Gjergji's proof IS correct. One only has to check that $\sum_{i,j} x_iz_j \otimes_R y_iw_j$ does indeed satisfy $b\sum_{i,j} x_iz_j \otimes_R y_iw_j=\sum_{i,j} x_iz_j \otimes_R y_iw_jb$ for every $b\in B$. This is best done by defining a map $A\otimes_B A\to A\otimes_R A,\ u\otimes_B v\mapsto f\cdot\left(u\otimes_R v\right)$, where $f$ is the separation element for $R\to B$ sent to $A\otimes_R A$ via the canonical homomorphism $B\to A$. One has to check that this is well-defined, but this is easy. – darij grinberg Feb 7 '11 at 11:49

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