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separable algebra

The idea

The concept of separable algebra is a strengthening of the concept of semisimple algebra, and a generalization of the concept of a separable field extension.

Definition

There are a several equivalent characterizations of separable algebras. For all of these we fix a field $k$. In what follows, all $k$-algebras will be assumed associative and unital.

First, a $k$-algebra $A$ is defined to be separable if for every field extension $K$ of $k$, the algebra $A \otimes_k K$ is semisimple.

Second, a $k$-algebra $A$ is separable if and only if it is flat when considered as a right module of $A^e = A \otimes_k A^{op}$ in the obvious (but perhaps not quite standard) way.

Third, a $k$-algebra $A$ is separable if and only if it is projective when considered as a left module of $A^e$ in the usual way.

Fourth, a $k$-algebra $A$ is separable if and only if the $A^e$-module morphism

$$\array{ m : & A^e &\to & A \\ & a \otimes b & \mapsto & a b }$$

has a right inverse, that is a $A^e$-module morphism

$$f: A \to A^e$$

with $m f = 1_A$.

It is easy to see that the third and fourth definitions are equivalent. We have an epimorphism of $A^e$-modules

$$A^e \stackrel{m}{\longrightarrow} A \to 0$$

If $f$ as above exists, this splits, so $A$ is a summand of a free $A^e$-module, namely $A^e$ itself, so $A$ is projective as an $A^e$-module. Conversely, if $A$ is projective, any epimorphism to $A$ splits.

We can also state the fourth characterization in a more grungy way in terms of the element $p = f(1)$. Namely, a $k$-algebra $A$ is separable if and only if there exists an element

$$p = \sum_{i=1}^n a_i \otimes b_i \in A^e$$

such that

$$\sum_{i=1}^n a_i b_i = 1$$

and

$$a p = p a$$

for all $a \in A$. Such an element $p$ is called a separability idempotent, since it satisfies $p^2 = p$. While grungy, finding a separability idempotent is a practical way to prove that an algebra is separable.

Classification

There is a classification theorem for separable algebras: separable algebras are the same as finite products of matrix algebras over finite-dimensional division algebras whose centers are finite-dimensional separable field extensions of the field $k$.

A perfect field is one for which every extension of is separable. Examples include fields of characteristic zero, or finite fields, or algebraically closed fields, or extensions of perfect fields. If $k$ is a perfect field, separable algebras are the same as finite products of matrix algebras over finite-dimensional division algebras over field $k$. In other words, if $k$ is a perfect field, there is no difference between a separable algebra over $k$ and a finite-dimensional semisimple algebra over $k$.

Relation to Frobenius algebras

A result of Eilenberg and Nakayama that any separable algebra over a field $k$ can be given the structure of a symmetric Frobenius algebra. Since the underlying vector space of a Frobenius algebra is isomorphic to its dual, any Frobenius algebra is necessarily finite dimensional, and so the same is true for separable algebras. For more details, see:

• Samuel Eilenberg and Tadasi Nakayama, On the dimension of modules and algebras. II. Frobenius algebras and quasi-Frobenius rings, Nagoya Math. J. 9 (1955), 1-16. (web)

A separable algebra is said to be strongly separable if there exists a separability idempotent $p$ that is symmetric, meaning

$$p = \sum_{i=1}^n a_i \otimes b_i = \sum_{i=1}^n b_i \otimes a_i$$

An algebra is strongly separable if and only if it can be made into a special Frobenius algebra. When this can be done, it can be done in a unique way.

There is an equivalent characterization of strongly separable algebras which makes this fact clearer. Any element $a$ of an associative unital algebra gives a left multiplication map

$$\array{ L_a : &A &\to& A \\ &b &\mapsto& a b }$$

When $A$ is finite-dimensional, there is a bilinear pairing $g: A \times A \to k$ defined by

$$g(a,b) = tr(L_a L_b)$$

An algebra $A$ is strongly separable if and only if $g$ is nondegenerate, i.e., if there is an isomorphism $A \to A^*$ given by

$$a \mapsto g(a, -)$$

In this case, there is just one way to make $A$ into a special Frobenius algebra, namely by defining the counit to be

$$\epsilon(a) = tr(L_a)$$

Here are some examples of strongly separable algebras:

• the algebra of $n \times n$ matrices with entries in the field $k$ is strongly separable if and only if $n$ is not divisible by the characteristic of $k$.

• the group algebra $k[G]$ of a finite group is strongly separable if and only if the order of $G$ is not divisible by the characteristic of $k$.

For more details, see Aguiar’s book below.

Over commutative rings

More generally, if $k$ is any unital commutative ring, we can define a separable $k$-algebra to be an algebra $A$ such that $A$ is projective as a module over $A^e = A \otimes_k A^{op}$.

As in the case of algebras over a field, an algebra $A$ over a commutative ring $k$ is separable if and only if the $A^e$-module epimorphism

$$\array{ m : & A^e &\to & A \\ & a \otimes b & \mapsto & a b }$$

splits, and this in turn is equivalent to the existence of a separability idempotent.

If a separable algebra $A$ is also projective as a module over $k$, it must be finitely generated as a $k$-module. For more details see DeMeyer-Ingraham.

Separable extensions of noncommutative rings

The ring extension $S$ over $R$ is said to be a separable extension if all short exact sequences of $S$-$S$-bimodules that are split as $S$-$R$-bimodules also split as $S$-$S$-bimodules. This is equivalent to the statement that the relative Hochschild cohomology $HH^n(S,R;M) = 0$ for all $n\gt 0$ and all coefficient bimodules $M$.

In algebraic geometry

Commutative separable algebras are important in algebraic geometry. The concept of étale cover in algebraic geometry is sort of a combination of covering space and separable algebra business. Lieven Le Bruyn has written “in categorical terms, studying the monoidal cat of commutative separable $k$-algebras is the same as studying the étale site of $k$”. This stuff would be nice to make precise…

Separable algebras play a major role in the Galois theory of extensions of algebras. Every separable $k$-algebra is a filtered colimit of finite-dimensional separable $k$-algebras???

There are further generalizations, leading to separable functors…

Literature and links

• $n$Lab: separable functor, separable field extension, separable coring

• wikipedia separable algebra

• Marcelo Aguiar, A note on strongly separable algebras, Boletín de la Academia Nacional de Ciencias (Córdoba, Argentina), special issue in honor of Orlando Villamayor, 65 (2000) 51-60. (web)

• F. DeMeyer and E. Ingraham, Separable algebras over commutative rings, Lecture Notes in Mathematics 181, Springer, Berlin, 1971.

• K. Hirata, K. Sugano, On semisimple and separable extensions of noncommutative rings, J. Math. Soc. Japan 18 (1966), 360-373.

An explicit proof of the Grothendieck Galois theory statement that the category of separable algebras over a field $K$ is anti-equivalent to the category of continuous actions on finite sets of the profinite fundamental group of $K$:

• Federico G. Lastaria, On separable algebras in Grothendieck Galois theory, Le Mathematiche 51:3, 1996, link