Posted by Jason Polak on 22. November 2016 · Write a comment · Categories: commutative-algebra · Tags: ,

Let $R$ be a commutative ring and $A$ be an $R$-algebra. We say that $A$ is a separable $R$-algebra if $A$ is projective as an $A\otimes_R A^{\rm op}$-module, where the action of $A\otimes_RA^{\rm op}$ is given by $(a\otimes a’)b = aba’$.

We already showed that the ring of upper triangular matrices over a commutative ring is not separable. This type of example is along the lines of: given a ring $R$, give some examples of $R$-algebras that are not separable.

On the other hand, you could also try constructing examples of non-separable algebras by taking a commutative ring $S$ as asking: what subrings $R$ of $S$ are such that $S$ is not a separable $R$-algebra? One way of producing subrings of $S$ is taking a finite group $G$ of automorphisms of $S$ and setting $R = S^G$. Then $S$ is an $R$-algebra and $G$ is a group of $R$-algebra automorphisms of $S$. Then:

Theorem. Suppose $S$ has no nontrivial idempotents. Then, $S$ is a separable $R = S^G$ algebra if and only if for each maximal ideal $M$ of $S$ and for each nontrivial $\sigma\in G$, there exists an $x\in S$ such that $\sigma(x) – x\not\in M$.

For example, if $k$ is a field then $k[x,y]$ is not separable over $k[x+y,xy]$, where the algebra structure is given via the inclusion map $k[x+y,xy]\to k[x,y]$.

Why? It’s easy to check that $k[x,y]^{\Z/2} = k[x+y,xy]$ where the nontrivial element $\sigma$ of $\Z/2$ permutes $x$ and $y$. Consider the maximal ideal $M = (x,y)$. For any $f\in k[x,y]$, the element $\sigma(f) – f$ is a polynomial without constant term, and hence is in the ideal $(x,y)$. Therefore, by the stated theorem, $k[x,y]$ cannot be separable as a $k[x+y,xy]$ algebra. Of course, there’s nothing special about $\Z/2$ here: this example works for any finite subgroup of permutations of the variables $x_1,\dots,x_n$ acting on $k[x_1,\dots,x_n]$.

I should also mention that $k[X] := k[x_1,\dots,x_n]$ is not separable as a $k$-algebra either, even though it is “classically separable” (in the sense that $K[X]$ has zero Jacobson radical for every extension $K/k$). That $k[X]$ is not separable in our sense follows because in general, if $A$ is both $R$-separable and $R$-projective, then it must also be finitely generated as an $R$-module.