Let be a commutative ring and let be a -algebra. In this post we’ll investigate a condition on which generalizes the condition that is a finite separable field extension (in the case that is a field). It can be stated in many equivalent ways, as follows. Below, “bimodule” always means “bimodule over .”

**Definition-Theorem:** The following conditions on are all equivalent, and all define what it means for to be a **separable **-algebra:

- is projective as an -bimodule (equivalently, as a left -module).
- The multiplication map has a section as an -bimodule map.
- admits a
**separability idempotent**: an element such that and for all (which implies that ).

(**Edit, 3/27/16: **Previously this definition included a condition involving Hochschild cohomology, but it’s debatable whether what I had in mind is the correct definition of Hochschild cohomology unless is a field or is projective over . It’s been removed since it plays no role in the post anyway.)

When is a field, this condition turns out to be a natural strengthening of the condition that is semisimple. In general, loosely speaking, a separable -algebra is like a “bundle of semisimple algebras” over .

**Proofs that the above conditions are equivalent**

: the multiplication map

is an epimorphism of -bimodules, so if is projective as an -bimodule then it splits, meaning it has a section. Conversely, since is a free -bimodule, if this map has a section then is a retract of a free -bimodule, hence is projective.

: a section of the multiplication map, as above, is determined by what it does to ; let’s call the image , and abuse terminology by identifying with the section it defines. What does it mean for to split the multiplication map? As a splitting, it must satisfy

since it’s the image of . Second, as an -bimodule map, it must satisfy for all , since in . (In fact is the free -bimodule on a generator with this property.) These conditions together imply that

hence that is an idempotent, as the name “separability idempotent” suggests. Hence a splitting of the multiplication map is the same data as a separability idempotent, which is in fact an idempotent.

This concludes the proof.

**A note on **

Above we chose to write the source of the multiplication map as to emphasize that it is the free -module, or equivalently the free -bimodule, on a generator. However, it can just as well be written , provided that we remember that the natural -bimodule structure on this is given by left multiplication on the first copy of and right multiplication on the second copy of . (Also, when we think of the separability idempotent as an idempotent, we really have in mind the algebra structure on .) This is how we’ll write things down in examples below.

**Some examples**

*Example.* itself is a separable -algebra, since it is even free as a -module.

*Example.* The matrix algebra is a separable -algebra. We can prove this explicitly by writing down a separability idempotent. Letting be the usual basis of (with relations , and all other multiplications zero), set

where is fixed. We have

and

while

so is a separability idempotent.

*Example.* If is a finite group whose order is invertible in , then the group algebra is a separable -algebra. Again we can prove this explicitly by writing down a separability idempotent. Set

.

We have

and, for any ,

where in the second line we made the substitution . So is a separability idempotent.

To get some more examples it will be convenient to use the following lemma.

**Lemma:** If is semisimple, then is separable over .

*Proof.* A ring is semisimple iff every module over it is projective. So if is semisimple, then in particular is a projective -module.

**Corollary:** If is a finite separable extension of a field (in the usual sense), then is separable over (in the above sense).

*Proof.* By the primitive element theorem, for some irreducible separable polynomial . Hence

where are the irreducible factors of over (it has at least two, since by definition contains a root of ). This is a finite product of fields and hence semisimple, so by the lemma we conclude that is separable.

Over a field, this gives another way to prove that the matrix algebras and the group algebras (where is invertible in ) are separable: is semisimple, and so is . But the proofs via writing down separability idempotents work for much more general base rings .

**Some general lemmas**

**Lemma:** is a separable -algebra iff is.

*Proof.* The opposite of a separability idempotent for is a separability idempotent for .

More explicitly, suppose is a separability idempotent for . Then is a separability idempotent for .

**Lemma:** If and are separable -algebras, then so is .

*Proof.* The sum of separability idempotents for and is a separability idempotent for .

More explicitly, recall that tensor product distributes over finite products for algebras. (In the commutative case this means that product distributes over finite coproducts for affine schemes.) Hence

and this is even an isomorphism of -bimodules, respecting the multiplication map down to . From this it’s not hard to verify that if is a separability idempotent for , and is a separability idempotent for , then , included into the above, is a separability idempotent for .

**Lemma:** If and are separable -algebras, then so is .

*Proof.* The tensor product of separability idempotents for and is a separability idempotent for .

**Lemma:** If is a separable -algebra, then the base change is a separable -algebra, for any commutative -algebra . (Hence separability is a **geometric** property in the strong sense that it is preserved by arbitrary base change.)

*Proof.* A separability idempotent for remains a separability idempotent for .

Note that this is not true for semisimple algebras, since an inseparable extension of the ground field is a counterexample.

**Lemma:** Any quotient of a separable algebra is separable.

*Proof.* A separability idempotent for remains a separability idempotent for any quotient of .

**Corollary: **Two -algebras are separable over if and only if is separable over .

*Proof.* In one direction, if are separable, then so is . In the other, if is separable, then are quotients of it, hence are also separable.

**Lemma:** Separability is Morita invariant: if and are Morita equivalent over , then is separable over iff is.

*Proof.* By the Eilenberg-Watts theorem, the category of -bimodules is equivalent (even monoidally) to the category of cocontinuous -linear endofunctors of . Among these, the bimodule itself represents the identity functor. Hence separability is equivalent to the condition that the identity is projective, and since this condition can be stated entirely in terms of it is Morita invariant.

**Lemma:** If is a commutative -algebra and is an -algebra such that 1) is separable over and 2) is separable over , then is separable over .

*Proof.* By hypothesis, the multiplication maps and split as bimodule maps, and we want to know that the same is true of the multiplication map . (Note that we are writing even though is commutative and so canonically isomorphic to its opposite; we don’t want to use this isomorphism.) If we write

then we can factor the multiplication map as a composite of two maps we know how to split, namely

where the first map applies the multiplication map between the two copies of and the second map is the multiplication map . (A similar argument can be used to show that the tensor product of separable algebras is separable.)

**Classification over a field**

We now classify separable -algebras when is a field.

**Lemma:** If is separable over a field , then is semisimple.

*Proof.* Any -bimodule describes a cocontinuous -linear functor as follows:

.

The bimodule represents the identity functor, while the free bimodule represents the functor

.

Consequently, the multiplication map represents the natural transformation

given by the action of . Now, if is separable, then this natural transformation splits, and in particular all of these action maps split. If is a field, then is a free -module, so is a free -module. This means that every -module is a retract of a free -module, hence is projective, and so is semisimple as desired.

**Corollary:** If is separable over a field , then is a finite product of matrix algebras over division algebras over , all of which must also be separable.

*Proof.* Since is semisimple, Artin-Wedderburn implies that for some division rings over . The lemmas we proved above imply that is separable iff are separable and that is separable iff is separable, so any such product is separable iff each is separable.

**Corollary:** If is separable over a field , then is **geometrically semisimple**: is semisimple for every field extension .

*Proof.* We know that separability is geometric (preserved by base change), so is separable over for every . If is a field extension, then the above lemma implies that is also semisimple.

(**Edit, 2/6/18:** Previously this section contained an incorrect proof that a separable algebra is finite-dimensional, as pointed out by Antoine in the comments.)

**Corollary:** An algebra over a field is separable over iff is semisimple.

*Proof.* Above we showed that if is semisimple, then is separable over (since every module, and in particular , is projective over ). Conversely, we also showed that tensor products and opposites of separable algebras are separable, so if is separable over then so is , and so it must also be semisimple.

At this point we’ve reduced to looking for division algebras over such that is semisimple. We’ll find them by inspecting their centers , which are field extensions of .

**Lemma:** Let be algebras over a field , and let denote the center. Then .

*Proof.* Suppose is central. This is equivalent to the condition that

for all and

.

for all . Since we’re working over a field, we can assume WLOG that the and are linearly independent in and respectively, from which it follows that these conditions hold if and only if for all . Hence , which (again, since we’re working over a field) is naturally a subalgebra of , and so must be the entire thing.

**Lemma:** If is a separable algebra over a field , then so is its center .

*Proof.* We know that is separable iff is semisimple. By the above lemma, we have

and since the center of a semisimple algebra is (a finite product of fields, hence) semisimple, it follows that is semisimple, hence (by another lemma) that is separable.

**Theorem:** Let be a field extension of a field . The following conditions are equivalent:

- is a finite separable extension of (in the usual sense).
- is a separable -algebra (in the above sense).
- is geometrically semisimple: is semisimple for all field extensions .
- is a finite product of copies of .

(**Edit, 2/7/18:** The statement of this result has been modified to be correct but, because of the error above, the proof is incomplete. In particular the proofs of and below are incomplete. I don’t know when I’ll get around to completing them, unfortunately.)

*Proof.* : we proved this above from the primitive element theorem.

: follows from a lemma above.

: set .

: any generates a -subalgebra of isomorphic to where is the minimal polynomial of . If is semisimple, it is a finite product of copies of , hence so is every -subalgebra of it. And is a finite product of copies of iff is separable over .

**Corollary:** If a division algebra over a field is separable over , then is finite-dimensional over , and its center is a finite separable extension of (in the usual sense).

This necessary condition in fact turns out to be sufficient. We need one more lemma to prove this, which is the following.

**Theorem:** Let be a **central simple algebra** over a field : that is, is a finite-dimensional simple -algebra with center . Then , where .

*Proof.* naturally acts on ( acting from the left, acting from the right). Simplicity of means that has no nontrivial two-sided ideals, or equivalently has no nontrivial -submodules, hence is simple as an -module. Its endomorphism ring is the center , and as a module over this endomorphism ring, . Hence the natural action of on gives a map

and we want this map to be a bijection. But it is a surjection by the Jacobson density theorem, hence a bijection since both sides have dimension over .

**Corollary:** A central simple algebra over a field is separable over .

**Corollary:** A finite-dimensional division algebra over a field whose center is a finite separable extension of is separable over .

*Proof.* We now know that is separable over , and that is separable over . By a lemma, it follows that is separable over .

**Corollary:** The separable algebras over a field are precisely the finite products of matrix algebras over finite-dimensional division algebras over whose centers are separable extensions of .

Over a perfect field, the last condition is automatic, so this just says that the separable algebras over are precisely the finite-dimensional semisimple -algebras.

on February 23, 2019 at 5:45 pm |KursatI think in the matrix algebra case you need to divide separability idempotent by n. For example, take $i=1$ and consider $e_{11}$ in the multiplication. For $n$ different $j$ values you get $n$ many $e_{11}$. In particular, $n$ must be divisible in $k$.

on March 22, 2019 at 12:43 pm |Qiaochu YuanThanks for the comment! is fixed and I’m not summing over it.

on February 1, 2018 at 8:41 am |AntoineI don’t understand the sentence “(since there are no nontrivial division algebras over an algebraically closed field)” in one of the corollaries of the classification of separable algebras.

I am not at all an expert of this (this is why I read your blog) but I believe there are nontrivial division algebras over algebraically closed fields, but they must be infinite dimensional.

More simply, if k is algebraically closed, A=k(t) gives a semi-simple k-algebra which is not finite dimensional over k.

Am I making a mistake ?

on February 3, 2018 at 5:58 pm |Qiaochu YuanYou’re right. It looks like the argument you’re quoting needs to be repaired; is not geometrically semisimple because it’s not semisimple after base change to , but I need an additional argument to establish this sort of thing.