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## Separable algebras

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

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

1. $A$ is projective as an $(A, A)$-bimodule (equivalently, as a left $A \otimes_k A^{op}$-module).
2. The multiplication map $A \otimes_k A^{op} \ni (a, b) \xrightarrow{m} ab \in A$ has a section as an $(A, A)$-bimodule map.
3. $A$ admits a separability idempotent: an element $p \in A \otimes_k A^{op}$ such that $m(p) = 1$ and $ap = pa$ for all $a \in A$ (which implies that $p^2 = p$).

(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 $k$ is a field or $A$ is projective over $k$. It’s been removed since it plays no role in the post anyway.)

When $k$ is a field, this condition turns out to be a natural strengthening of the condition that $A$ is semisimple. In general, loosely speaking, a separable $k$-algebra is like a “bundle of semisimple algebras” over $\text{Spec } k$.

Proofs that the above conditions are equivalent

$1 \Leftrightarrow 2$: the multiplication map

$\displaystyle A \otimes_k A^{op} \ni a \otimes b \xrightarrow{m} ab \in A$

is an epimorphism of $(A, A)$-bimodules, so if $A$ is projective as an $(A, A)$-bimodule then it splits, meaning it has a section. Conversely, since $A \otimes_k A^{op}$ is a free $(A, A)$-bimodule, if this map has a section then $A$ is a retract of a free $(A, A)$-bimodule, hence is projective.

$2 \Leftrightarrow 3$: a section of the multiplication map, as above, is determined by what it does to $1 \in A$; let’s call the image $p = \sum a_i \otimes b_i$, and abuse terminology by identifying $p$ with the section it defines. What does it mean for $p$ to split the multiplication map? As a splitting, it must satisfy

$\displaystyle m(p) = \sum a_i b_i = 1 \in A$

since it’s the image of $1$. Second, as an $(A, A)$-bimodule map, it must satisfy $ap = pa$ for all $a \in A$, since $a \cdot 1 = 1 \cdot a$ in $A$. (In fact $A$ is the free $(A, A)$-bimodule on a generator with this property.) These conditions together imply that

$\displaystyle p^2 = \sum_i a_i p b_i = \left( \sum_i a_i b_i \right) p = p$

hence that $p$ 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. $\Box$

A note on $A \otimes_k A^{op}$

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

Some examples

Example. $k$ itself is a separable $k$-algebra, since it is even free as a $k \otimes_k k^{op} \cong k$-module.

Example. The matrix algebra $M_n(k)$ is a separable $k$-algebra. We can prove this explicitly by writing down a separability idempotent. Letting $e_{ij}, 1 \le i, j \le n$ be the usual basis of $M_n(k)$ (with relations $e_{ij} e_{jk} = e_{ik}$, and all other multiplications zero), set

$\displaystyle p = \sum_i e_{ij} \otimes e_{ji} \in M_n(k) \otimes M_n(k)$

where $j$ is fixed. We have

$\displaystyle m(p) = \sum_i e_{ij} e_{ji} = \sum_i e_{ii} = 1$

and

$e_{ab} p = \sum_i e_{ab} e_{ij} \otimes e_{ji} = e_{aj} \otimes e_{jb}$

while

$\displaystyle p e_{ab} = \sum_i e_{ij} \otimes e_{ji} e_{ab} = e_{aj} \otimes e_{jb}$

so $p$ is a separability idempotent.

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

$\displaystyle p = \frac{1}{|G|} \sum_g g \otimes g^{-1} \in k[G] \otimes k[G]$.

We have

$\displaystyle m(p) = \frac{1}{|G|} \sum_g g g^{-1} = 1 \in k[G]$

and, for any $h \in G$,

\displaystyle \begin{aligned} hp &= \frac{1}{|G|} \sum_g hg \otimes g^{-1} \\ &= \frac{1}{|G|} \sum_g g \otimes (h^{-1} g)^{-1} \\ &= \frac{1}{|G|} \sum_g g \otimes g^{-1} h \\ &= ph \end{aligned}

where in the second line we made the substitution $hg \mapsto g$. So $p$ is a separability idempotent.

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

Lemma: If $A \otimes_k A^{op}$ is semisimple, then $A$ is separable over $k$.

Proof. A ring is semisimple iff every module over it is projective. So if $A \otimes_k A^{op}$ is semisimple, then in particular $A$ is a projective $A \otimes_k A^{op}$-module. $\Box$

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

Proof. By the primitive element theorem, $L = k[x]/f(x)$ for some irreducible separable polynomial $f(x)$. Hence

$\displaystyle L \otimes_k L^{op} \cong L[x]/f(x) \cong \prod_i L[x]/f_i(x)$

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

Over a field, this gives another way to prove that the matrix algebras $M_n(k)$ and the group algebras $k[G]$ (where $|G|$ is invertible in $k$) are separable: $M_n(k) \otimes_k M_n(k)^{op} \cong M_{n^2}(k)$ is semisimple, and so is $k[G] \otimes_k k[G]^{op} \cong k[G \times G^{op}]$. But the proofs via writing down separability idempotents work for much more general base rings $k$.

Some general lemmas

Lemma: $A$ is a separable $k$-algebra iff $A^{op}$ is.

Proof. The opposite of a separability idempotent for $A$ is a separability idempotent for $A^{op}$.

More explicitly, suppose $p = \sum a_i \otimes b_i \in A \otimes_k A^{op}$ is a separability idempotent for $A$. Then $p^{op} = \sum b_i \otimes a_i \in A^{op} \otimes A$ is a separability idempotent for $A^{op}$. $\Box$

Lemma: If $A$ and $B$ are separable $k$-algebras, then so is $A \times B$.

Proof. The sum of separability idempotents for $A$ and $B$ is a separability idempotent for $A \times B$.

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

$(A \times B) \otimes_k (A \times B) \cong (A \otimes_k A) \times (A \otimes_k B) \times (B \otimes_k A) \times (B \otimes_k B)$

and this is even an isomorphism of $(A \times B, A \times B)$-bimodules, respecting the multiplication map down to $A \times B$. From this it’s not hard to verify that if $p_A \in A \otimes_k A$ is a separability idempotent for $B$, and $p_B \in B \otimes_k B$ is a separability idempotent for $B$, then $p_A + p_B \in (A \otimes_k A) \times (B \otimes_k B)$, included into the above, is a separability idempotent for $A \times B$$\Box$

Lemma: If $A$ and $B$ are separable $k$-algebras, then so is $A \otimes_k B$.

Proof. The tensor product of separability idempotents for $A$ and $B$ is a separability idempotent for $A \otimes_k B$. $\Box$

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

Proof. A separability idempotent for $A$ remains a separability idempotent for $A \otimes_k K$. $\Box$

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 $A$ remains a separability idempotent for any quotient of $A$. $\Box$

Corollary: Two $k$-algebras $A, B$ are separable over $k$ if and only if $A \times B$ is separable over $k$.

Proof. In one direction, if $A, B$ are separable, then so is $A \times B$. In the other, if $A \times B$ is separable, then $A, B$ are quotients of it, hence are also separable. $\Box$

Lemma: Separability is Morita invariant: if $A$ and $B$ are Morita equivalent over $k$, then $A$ is separable over $k$ iff $B$ is.

Proof. By the Eilenberg-Watts theorem, the category of $(A, A)$-bimodules is equivalent (even monoidally) to the category of cocontinuous $k$-linear endofunctors of $\text{Mod}(A)$. Among these, the bimodule $A$ 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 $\text{Mod}(A)$ it is Morita invariant. $\Box$

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

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

$\displaystyle A \otimes_k A^{op} \cong A \otimes_R (R \otimes_k R^{op}) \otimes_{R^{op}} A^{op}$

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

$\displaystyle A \otimes_R (R \otimes_k R^{op}) \otimes_{R^{op}} A^{op} \to A \otimes_R A^{op} \to A$

where the first map applies the multiplication map $R \otimes_k R^{op} \to R$ between the two copies of $A$ and the second map is the multiplication map $A \otimes_R A^{op} \to A$. (A similar argument can be used to show that the tensor product of separable algebras is separable.) $\Box$

Classification over a field

We now classify separable $k$-algebras when $k$ is a field.

Lemma: If $A$ is separable over a field $k$, then $A$ is semisimple.

Proof. Any $(A, A)$-bimodule $M$ describes a cocontinuous $k$-linear functor $\text{Mod}(A) \to \text{Mod}(A)$ as follows:

$\displaystyle \text{Mod}(A) \ni V \mapsto M \otimes_A V \in \text{Mod}(A)$.

The bimodule $A$ represents the identity functor, while the free bimodule $A \otimes_k A$ represents the functor

$\displaystyle \text{Mod}(A) \ni V \mapsto A \otimes_k V \ni \text{Mod}(A)$.

Consequently, the multiplication map $A \otimes_k A \xrightarrow{m} A$ represents the natural transformation

$\displaystyle A \otimes_k V \to V$

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

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

Proof. Since $A$ is semisimple, Artin-Wedderburn implies that $A \cong \prod_{i=1}^n M_{d_i}(D_i)$ for some division rings $D_i$ over $k$. The lemmas we proved above imply that $A \times B$ is separable iff $A, B$ are separable and that $A$ is separable iff $M_d(A)$ is separable, so any such product is separable iff each $D_i$ is separable. $\Box$

Corollary: If $A$ is separable over a field $k$, then $A$ is geometrically semisimple$A \otimes_k L$ is semisimple for every field extension $k \to L$.

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

(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 $A$ over a field $k$ is separable over $k$ iff $A \otimes_k A^{op}$ is semisimple.

Proof. Above we showed that if $A \otimes_k A^{op}$ is semisimple, then $A$ is separable over $k$ (since every module, and in particular $A$, is projective over $A \otimes_k A^{op}$). Conversely, we also showed that tensor products and opposites of separable algebras are separable, so if $A$ is separable over $k$ then so is $A \otimes_k A^{op}$, and so it must also be semisimple. $\Box$

At this point we’ve reduced to looking for division algebras $D$ over $k$ such that $D \otimes_k D^{op}$ is semisimple. We’ll find them by inspecting their centers $Z(D)$, which are field extensions of $k$.

Lemma: Let $A, B$ be algebras over a field $k$, and let $Z(-)$ denote the center. Then $Z(A \otimes_k B) \cong Z(A) \otimes_k Z(B)$.

Proof. Suppose $z = \sum a_i \otimes b_i \in Z(A \otimes_k B)$ is central. This is equivalent to the condition that

$\displaystyle [z, a \otimes 1] = \sum [a_i, a] \otimes b_i = 0$

for all $a \in A$ and

$\displaystyle [z, 1 \otimes b] = \sum a \otimes [b_i, b] = 0$.

for all $b \in B$. Since we’re working over a field, we can assume WLOG that the $a_i$ and $b_i$ are linearly independent in $A$ and $B$ respectively, from which it follows that these conditions hold if and only if $[a_i, a] = [b_i, b] = 0$ for all $i, a, b$. Hence $z \in Z(A) \otimes_k Z(B)$, which (again, since we’re working over a field) is naturally a subalgebra of $Z(A \otimes_k B)$, and so must be the entire thing. $\Box$

Lemma: If $A$ is a separable algebra over a field $k$, then so is its center $Z(A)$.

Proof. We know that $A$ is separable iff $A \otimes_k A^{op}$ is semisimple. By the above lemma, we have

$\displaystyle Z(A \otimes_k A^{op}) \cong Z(A) \otimes_k Z(A)^{op}$

and since the center of a semisimple algebra is (a finite product of fields, hence) semisimple, it follows that $Z(A) \otimes_k Z(A)^{op}$ is semisimple, hence (by another lemma) that $Z(A)$ is separable. $\Box$

Theorem: Let $L$ be a field extension of a field $k$. The following conditions are equivalent:

1. $L$ is a finite separable extension of $k$ (in the usual sense).
2. $L$ is a separable $k$-algebra (in the above sense).
3. $L$ is geometrically semisimple: $L \otimes_k L'$ is semisimple for all field extensions $k \to L'$.
4. $L \otimes_k \bar{k}$ is a finite product of copies of $\bar{k}$.

(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 $3 \Rightarrow 4$ and $4 \Rightarrow 1$ below are incomplete. I don’t know when I’ll get around to completing them, unfortunately.)

Proof. $1 \Rightarrow 2$: we proved this above from the primitive element theorem.

$2 \Rightarrow 3$: follows from a lemma above.

$3 \Rightarrow 4$: set $L' = \bar{k}$.

$4 \Rightarrow 1$: any $\alpha \in L$ generates a $\bar{k}$-subalgebra of $L \otimes_k \bar{k}$ isomorphic to $\bar{k}[\alpha]/f(\alpha)$ where $f$ is the minimal polynomial of $\alpha$. If $L \otimes_k \bar{k}$ is semisimple, it is a finite product of copies of $\bar{k}$, hence so is every $\bar{k}$-subalgebra of it. And $\bar{k}[\alpha]/f(\alpha)$ is a finite product of copies of $\bar{k}$ iff $f$ is separable over $k$. $\Box$

Corollary: If a division algebra $D$ over a field $k$ is separable over $k$, then $D$ is finite-dimensional over $k$, and its center $Z(D)$ is a finite separable extension of $k$ (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 $A$ be a central simple algebra over a field $L$: that is, $A$ is a finite-dimensional simple $L$-algebra with center $L$. Then $A \otimes_L A^{op} \cong M_n(L)$, where $n = \dim_L A$.

Proof. $A \otimes_L A^{op}$ naturally acts on $A$ ($A$ acting from the left, $A^{op}$ acting from the right). Simplicity of $A$ means that $A$ has no nontrivial two-sided ideals, or equivalently has no nontrivial $A \otimes_L A^{op}$-submodules, hence $A$ is simple as an $A \otimes_L A^{op}$-module. Its endomorphism ring is the center $Z(A) \cong L$, and as a module over this endomorphism ring, $A \cong L^n$. Hence the natural action of $A \otimes_L A^{op}$ on $A$ gives a map

$\displaystyle A \otimes_L A^{op} \to \text{End}_L(A) \cong M_n(L)$

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 $n^2$ over $L$. $\Box$

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

Corollary: A finite-dimensional division algebra $D$ over a field $k$ whose center $Z(D)$ is a finite separable extension of $k$ is separable over $k$.

Proof. We now know that $D$ is separable over $Z(D)$, and that $Z(D)$ is separable over $k$. By a lemma, it follows that $D$ is separable over $k$. $\Box$

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

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

### 4 Responses

1. I 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 $k$.

• Thanks for the comment! $j$ is fixed and I’m not summing over it.

2. I 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 ?

• You’re right. It looks like the argument you’re quoting needs to be repaired; $k(t)$ is not geometrically semisimple because it’s not semisimple after base change to $k(t)$, but I need an additional argument to establish this sort of thing.