### Grothendieck–Galois–Brauer Theory (Part 6)

#### Posted by John Baez

I’ve been talking about Grothendieck’s approach to Galois theory, but I haven’t yet touched on Brauer theory. Both of these involve separable algebras, but of different kinds. For Galois theory we need *commutative* separable algebras, which are morally like covering spaces. For Brauer theory we’ll need separable algebras that are as *noncommutative as possible*, which are morally like bundles of matrix algebras. One of my ultimate goals is to unify these theories — or, just as likely, learn how someone has already done it, and explain what they did.

Both subjects are very general and conceptual. But to make sure I understand the basics, my posts so far have focused on the most classical case: separable algebras over fields. I’ve explained a few different viewpoints on them. It’s about time to move on. But before I do, I should at least *classify* separable algebras over fields.

In all that follows, $k$ is any field, and all algebras are over $k$. I’ll write $M_n(A)$ for the algebra of $n \times n$ matrices with entries in the algebra $A$. I’ll call this a **matrix algebra over $A$**.

Here’s the classification of separable algebras: *they’re all finite products of matrix algebras over separable division algebras*.

A division algebra is just an algebra where every nonzero element has a multiplicative inverse. A separable division algebra is just a division algebra that’s separable.

For example: you’ve probably heard that the real numbers, complex numbers and quaternions are division algebras over $k = \mathbb{R}$. These are the only *separable* division algebras over $\mathbb{R}$. There are lots of *other* division algebras over $\mathbb{R}$, but they’re all infinite-dimensional. In fact, over any field, separable division algebras must be finite-dimensional — and thus by the classification theorem, all separable algebras are finite-dimensional! But I will not prove this today.

Here’s that classification again:

**Theorem 21.** Suppose $D_i$ are separable division algebras. Then any finite product of matrix algebras $M_{n_i}(D_i)$ is separable. Conversely, any separable algebra is of this form.

This looks a lot like the Wedderburn–Artin theorem, which classifies semisimple algebras, and indeed we’ll reduce it to that. So I should remind you of that!

Remember, an object $A$ in an abelian category is **simple** if it has no nontrivial quotients — or equivalently, no nontrivial subobjects. An object in an abelian category is **semisimple** if it’s a (possibly infinite) direct sum of simple objects. An abelian category is **semisimple** if all its objects are semisimple. And finally — whew! — an algebra $A$ over $k$ is **semisimple** if the category $_{A}\mathrm{Mod}$ of left $A$-modules is semisimple.

We then have:

**Theorem 22 (Wedderburn–Artin Theorem).** Suppose $D_i$ are division algebras. Then any finite product of matrix algebras $M_{n_i}(D_i)$ is semisimple. Conversely, any semisimple algebra is of this form.

This is well-known, but the proof is so nice I’ll include it at the end of this article.

Using this, we can prove Theorem 21 if we can show three things:

**Lemma 23.** A product of algebras $A \times B$ is separable iff $A$ and $B$ are separable.

**Lemma 24.** For any algebra $A$, the matrix algebra $M_n(A)$ is separable iff $A$ is separable.

**Lemma 25.** Any separable algebra is semisimple.

The argument is then easy as pie. Suppose $D_i$ are separable division algebras. Then the matrix algebras $M_{n_i}(D_i)$ is separable by Lemma 24, so the product

$\prod_{i = 1}^m M_{n_i}(D_i)$

is separable by Lemma 23.

Conversely, suppose $A$ is a separable algebra. Then $A$ is semisimple by Lemma 25, so the Wedderburn–Artin theorem implies

$A \cong \prod_{i = 1}^m M_{n_i}(D_i)$

for some division algebras $D_i$. By Lemmas 23 and 24, the separability of $A$ implies the separability of each division algebra $D_i$. █

So, let’s prove the lemmas! This will be a pleasant romp through various ideas in algebra. My exposition today relies heavily on this wonderful blog post:

- Qiaochu Yuan, Separable algebras,
*Annoying Precision*, March 27, 2016.

If you want to learn more about separable algebras you should definitely read this!

### New separable algebras from old: separability idempotents

We’ll begin by proving Lemma 23, which says $A \times B$ is separable iff $A$ and $B$ are. For this it’s nice to think about separability in a very concrete and practical way. The idea is that $A$ is separable iff $A \otimes A$ contains an element $\pi$ with some special properties, called a ‘separability idempotent’. It’s called ‘idempotent’ because $\pi^2 = \pi$ if we think of it as an element of the algebra $A \otimes A^{\text{op}}$, but we won’t need that here.

First remember Lemma 16: an algebra $A$ is separable iff the exact sequence of $A,A$-bimodules

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

splits, where $m$ is the multiplication map, and we make $A \otimes A$ into an $A,A$-bimodule like this:

$a(b \otimes c) = a b \otimes c, (b \otimes c)a = b \otimes c a$

The splitting gives a map

$\Delta \colon A \to A \otimes A$

and since this is a map of bimodules and $A$ is generated as a bimodule by $1 \in A$, $\Delta$ is determined by the image of $1$, say

$\pi = \Delta(1) \in A \otimes A$

This element $\pi$ has two important properties. Since $\Delta$ is a bimodule map we have

$a \pi = a \Delta(1) = \Delta(a) = \Delta(1) a = \pi a$

and since $\Delta$ is a splitting we have

$m(\pi) = m(\Delta(1)) = 1$

So, we’ve shown half of this without even breaking a sweat:

**Lemma 26.** An algebra $A$ is separable iff there exists a **separability idempotent** for $A$: that is, $\pi \in A \otimes A$ with $m(\pi) = 1 \in A$ and $a \pi = \pi a$ for all $a \in A$.

The converse is just as easy: given $\pi$ obeying these equations we just define

$\Delta(a) = a \pi$

and check using our two equations that it’s a map of bimodules and a splitting.

Using this trick, we easily see:

**Lemma 27.** Any quotient of a separable algebra $A$ is separable.

**Proof.** Given a surjection $j \colon A \to B$ and a separability idempotent for $A$, we get a separability idempotent $(j \otimes j)\pi$ for the quotient algebra $B$. █

And now we’re ready to get the job done:

**Lemma 23.** A product $A \times B$ of algebras over $k$ is separable iff $A$ and $B$ are separable.

**Proof.** If $A \times B$ is separable its quotients $A$ and $B$ are separable by Lemma 27. Conversely, if we have separability idempotents $p$ for $A$ and $q$ for $B$ then $(p,q)$ is a separability idempotent for $A \times B$. █

It’s also easy to show *half* of Lemma 24 using this trick: if $A$ is separable we can pick a separability idempotent $\pi$ for it, and tensoring $\pi$ with an identity matrix we get a separability idempotent for $M_n(A)$, so $M_n(A)$ is also separable. But the converse is apparently not so easy using this method. It pays to turn to some more abstract ideas.

### Separable algebras and projective modules

We’ll use a beautiful connection between separable algebras and projective modules. There are various equivalent definitions of ‘projective module’, but we’ll just need two of the most famous:

a module is projective iff it is a summand of a free module,

a module $P$ is projective iff for any surjection of modules

$Y \stackrel{p}{\longrightarrow} X \longrightarrow 0$

we can lift any morphism $f \colon P \to X$ to a morphism $g \colon P \to Y$, i.e. we can find $g$ with $f = p \circ g$.

These results hold for bimodules, too, since an $A,B$-bimodule is the same as an $A \otimes B^{\text{op}}$-module.

Okay. Here is the beautiful connection:

**Lemma 28.** An algebra $A$ is separable iff it is projective as an $A,A$-bimodule.

**Proof.** The key is to use Lemma 16: an algebra $A$ is separable iff the exact sequence of $A,A$-bimodules

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

splits,

Now suppose $A$ is separable. Since $A \otimes A$ is a free $A,A$-bimodule and the above exact sequence splits, $A$ is a summand of a free bimodule, so it is projective.

Conversely, suppose $A$ is projective as an $A,A$-bimodule. The multiplication map $m \colon A \otimes A \to A$ is a surjection of bimodules, so we can lift the identity $1 \colon A \to A$ to a morphism $\Delta \colon A \to A \otimes A$. This means precisely that $\Delta$ splits the exact sequence of bimodules

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

so $A$ is separable. █

### New separable algebras from old: Morita equivalence

Morita equivalence will ultimately become a hugely important character in my story, but for now it’s just a tool, so I’ll treat it quite superficially here. Two algebras $A$ and $B$ over our field $k$ are **Morita equivalent** if their categories of left modules are equivalent as $k$-linear categories:

${}_A \mathrm{Mod} \simeq {}_B \mathrm{Mod}$

We say a property of algebras is **Morita invariant** if whenever $A$ has it and $B$ is Morita equivalent to $A$, then $B$ has it too. The intuition is that some property of an algebra is Morita invariant if you can detect it just by examining the category of modules of that algebra. And soon we’ll see that separability is Morita invariant!

One can show that any algebra $A$ is Morita equivalent to the matrix algebra $M_n(A)$: the equivalence sends any $A$-module $X$ to the $M_n(A)$-module $X^n$. Thus, to prove the next of our three big lemmas:

**Lemma 24.** For any algebra $A$, the matrix algebra $M_n(A)$ is separable iff $A$ is separable.

it suffices to show this:

**Lemma 29.** Separability is Morita invariant: if two algebras $A$ and $B$ are Morita equivalent, then $A$ is separable iff $B$ is separable.

**Proof.** By a version of the Eilenberg–Watts theorem, the category of $A,A$-bimodules is equivalent to the category of colimit-preserving $k$-linear functors

$F \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$

The idea is that tensoring with any $A,A$-bimodule gives such a functor $F$, and all such functors come from $A,A$-bimodules in this way (up to natural isomorphism). The $A,A$-bimodule $A$ gives the identity functor $1 \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$.

In Lemma 28 we showed that $A$ is separable iff $A$ is projective in the category of $A,A$-bimodules. Thus, $A$ is separable iff the identity functor $1 \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$ is projective in the category of colimit-preserving $k$-linear functors from ${}_A \mathrm{Mod}$ to itself. But this last characterization of separability depends only on ${}_A \mathrm{Mod}$. So, separability is Morita invariant! █

I got this beautiful proof from Qiaochu Yuan’s blog.

I used to be intimidated by the Eilenberg–Watts theorem, but then I realized it’s a spinoff of the Yoneda embedding for enriched categories. For ordinary categories, part of the Yoneda business is that the presheaf category $\widehat{C}$ of a small category $C$ is its ‘free cocompletion’. This implies that ‘bimodules’ from $C$ to $D$, also called profunctors, are equivalent to colimit-preserving functors $F \colon \widehat{C} \to \widehat{D}$. But if we work with categories enriched over $\mathrm{Vect}_k$ we get an analogous result. We can think of an algebra $A$ as a 1-object enriched category of this sort, and then its enriched presheaf category is ${}_A \mathrm{Mod}$, and $A,B$-bimodules are equivalent to colimit-preserving $k$-linear functors $F \colon {}_A \mathrm{Mod} \to {}_B \mathrm{Mod}$.

### Semisimple algebras and projective modules

We have just one more lemma to show: Lemma 25. This says that separable algebras are semisimple. To prove this we’ll use a nice connection between semisimple algebras and projective modules.

**Lemma 30.** An algebra $A$ is semisimple if every $A$-module is projective.

**Proof.** It’s easy to use the ‘lifting property’ definition of projectivity to show that if $P$ is a projective module, any short exact sequence

$0 \to M \to N \to P \to 0$

splits. Thus if every $A$-module is projective, every short exact sequence of $A$-modules splits. But an algebra $A$ is semisimple if every short exact sequence of $A$-modules splits. █

Most of the work is hidden in the last sentence! You can see a proof of that fact here. It’s a bit tiring and Zorny, so I won’t include it.

### Separable algebras are semisimple

Now we’re ready to prove our last big lemma:

**Lemma 25.** Any separable algebra is semisimple.

**Proof.** We use the Eilenberg-Watts theorem again: the category of $A,A$-bimodules is equivalent to the category of colimit-preserving $k$-linear functors

$F \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$

Tensoring with the $A$-bimodule $A$ gives the identity functor

$1 \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$

while tensoring with the $A$-bimodule $A \otimes A$ gives the functor

$A \otimes_k - \colon {}_A \mathrm{Mod} \to {}_A \mathrm{Mod}$

The multiplication map $m \colon A \otimes A \to A$ thus gives a natural transformation

$\alpha_M \otimes_k M \to M \qquad (M \in {}_A \mathrm{Mod})$

which is just the action of $A$ on $M$. If $A$ is separable the multiplication map splits, and thus all these action maps $\alpha_M$ split. Thus, each module $M$ is a summand of the module $A \otimes_k M$, which is a free module. Thus, every module of $A$ is projective. By Lemma 30, it follows that $A$ is semisimple. █

### The Wedderburn-Artin Theorem for algebras

**Theorem 22 (Wedderburn–Artin Theorem).** Suppose $D_i$ are division algebras. Then any finite product of matrix algebras $M_{n_i}(D_i)$ is semisimple. Conversely, any semisimple algebra is of this form.

**Proof.** If $D$ is a division algebra then every left $M_n(D)$-module is a direct sum of copies of the simple module $D^n$, so $D$ is semisimple. Also, if $A$ and $B$ are semisimple we can show $A \times B$ is as well, using the fact that $\mathrm{Mod}(A \times B) \simeq \mathrm{Mod}(A) \times \mathrm{Mod}(B)$. Thus any finite product of matrix algebras over division algebras is semisimple.

The interesting part is the other direction. Suppose $A$ is semisimple. Then by definition any left $A$-module is a direct sum of simple modules, and any finitely generated $A$-module must be a finite direct sum of such. $A$ is thus isomorphic as a left $A$-module to a finite direct sum of simple modules, which are minimal left ideals of $A$. Write this direct sum as

$A \;\cong\; \bigoplus_i I_i^{n_i}$

where the $I_i$ are mutually nonisomorphic simple $A$-modules, the $i$th one appearing with multiplicity $n_i$. Then the algebra of left $A$-module endomorphisms of $A$ is

$\mathrm{End}(A) \;\cong\; \prod_i \mathrm{End}\big(I^{n_i}\big)$

We can identify $\mathrm{End}\big(I_i^{n_i}\big)$ with an algebra of matrices

$\mathrm{End}\big(I_i^{n_i}\big) \;\cong\; M_{n_i}\big(\mathrm{End}(I_i)\big)$

But since $I_i$ is a simple module, the endomorphism algebra $\mathrm{End}(I_i)$ must be a division algebra, by Schur’s Lemma. (The proof of this lemma is instantaneous: if some nonzero element of this algebra had no inverse, either its kernel would be a nontrivial subobject of $I_i$, or its cokernel would be a nontrivial quotient object — both impossible because $I_i$ is simple.) Thus we have

$\mathrm{End}(A) \;\cong\; \prod_i M_{n_i}\big(D_i\big)$

for some division algebras $D_i$.

Since every endomorphism of $A$ as a left $A$-module is given by right multiplication, $\mathrm{End}(A) \cong A^{\text{op}}$ as algebras, and we get

$A^{\text{op}} \;\cong\; \prod_i M_{n_i}\big(D_i)$

But since a matrix algebra is isomorphic to its opposite (using the transpose of matrices), it follows that

$A \;\cong\; \prod_i M_{n_i}(D_i)$

as desired. █

By the way, while we defined an algebra to be semisimple if its category of *left* modules is semisimple, the Wedderburn–Artin theorem implies that an algebra is semisimple iff its category of *right* modules is semisimple. The reason is that right $A$-modules are equivalent to left $A^{\text{op}}$-modules, but we’ve just seen that a semisimple algebra has $A \cong A^{\text{op}}$.

Another way to put it is that an algebra $A$ is semisimple iff $A^{\text{op}}$ is semisimple. Similarly, our main theorem today implies that an algebra $A$ is separable iff $A^{\text{op}}$ is separable. So, these concepts are very different from Noetherianness, where we have distinct concepts of ‘left Noetherian’ and ‘right Noetherian’ rings, and $R$ is left Noetherian iff $R^{\text{op}}$ is right Noetherian.