What is the 'correct' definition of a separable algebra over a non-commutative ring? Are there known results about such algebras? Examples?

Recall that one of the equivalent definitions of a separable algebra $A$ over a commutative ring $R$ says that $A$ is a projective $A \otimes_R A^{op}$-module.

Also recall that there are several ways to define an algebra over a non-commutative ring, see the followimg two questions: 1 and 2.


1) An example for commutative $R$ and $A$ is: $R=k[p,q]$, $A=k[x,y]$, where $k$ is a field of characteristic zero and $p,q \in k[x,y]$ have an invertible Jacobian; see Theorem 38.

I wonder what can be said in the non-commutative analog, where $A$ is the first Weyl algebra (generated by $X$ and $Y$) and $R$ is its sub-algebra generated by the images of $X$ and $Y$ under an endomorphism.

2) If $R$ is non-commutative (and $A$ is non-commutative), then is there a problem with $A \otimes_R A^{op}$? Defining the tensor product of a left $R$-module $A$ with a right $R$-module $B$ over a non-commutative $R$ seems ok, but perhaps what we get is not a ring, but only a group? (I may be wrong).


1 Answer 1


The standard definition is: an inclusion of algebras $R\subseteq A$ is a separable extension of algebras if the map $\mu:A\otimes_RA\to A$ induced by the multiplication of $A$ is split as a map of $A$-bimodules.

This is used in many places and has many applications. For example, if $G$ is a finite group, $k$ a field of characteristic $p$ dividing the order of $G$ and let $P$ be a $p$-Sylow subgroup of $G$. Then the extension of group algebras $kP\subseteq kG$ is separable. This is important, for example, in proving Higman's theorem that says that $kG$ has finite representation type iff $P$ is cyclic, and in many other places.

You can read a bit about this in Pierce's book on associative algebras.


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