## June 14, 2023

### The Wedderburn–Artin Theorem

#### Posted by John Baez

Now that I finally understand the proof of the Wedderburn–Artin Theorem, I want to do what any good category theorist would do: formulate it more generally and abstractly, so nobody can understand me and I look like a genius everybody can see how simple it actually is.

The Wedderburn–Artin Theorem says that any semisimple ring $R$ is a finite product of matrix algebras over division rings. In symbols:

$\displaystyle{ R \cong \prod_{i = 1}^n M_{k_i}(D_i) }$

where $D_1, \dots , D_n$ are division rings.

But what’s a semisimple ring? We can define it to be one whose category of finitely generated modules is semisimple. Left or right modules? It turns out not to matter, though this is not obvious! So, pick one.

But what’s a semisimple category? It’s usually defined as an abelian category where every object is a finite direct sum of simple objects.

But what’s a simple object? It’s an object $X$ with no quotients other than $0$ and $X$ itself. In an abelian category, this equivalent to saying $X$ has no subobjects other than $0$ and $X$. We use both of these facts to see how division rings show up in the Wedderburn–Artin theorem. Here’s how:

Schur’s Lemma. If $X$ and $Y$ are simple objects in an abelian category, then any morphism $f \colon X \to Y$ is either an isomorphism or $0$. Thus, if $X$ is a simple object in an abelian category, $Hom(X,X)$ is a division ring.

Proof. If $X$ and $Y$ are simple the kernel of any morphism $f \colon X \to Y$ must be $0$ or $X$, and its cokernel must be $0$ or $Y$. If the kernel is $X$ or the cokernel is $Y$ we must have $f = 0$. The only alternative is that the kernel is $0$ and the cokernel is $0$. In an abelian category this implies that $f$ is an isomorphism. ▮

The proof of Wedderburn–Artin then goes as follows. Suppose $R$ is a ring whose category $Mod_R$ of finitely generated right modules is semisimple. Let $R_R \in Mod_R$ stand for $R$ viewed as a right module over itself. Then $R_R$ is a finite direct sum of nonisomorphic simple objects $X_1, \dots, X_k$, where each object $X_i$ can show up any number of times in the direct sum, say $n_i$ times:

$\displaystyle{ R_R \cong \bigoplus_{i = 1}^k X_i^{n_i} }$

Thus by Schur’s Lemma and the way direct sums work,

$\displaystyle{ End(R_R) \cong \bigoplus_{i = 1}^k M_{n_i}(End(X_i)) }$

so $End(R_R)$ is a finite product of matrix algebras over division rings!

Finally, we use the ring-theoretic analogue of Cayley’s Theorem: every endomorphism of $R$ as a right $R$-module is given by left multiplication by some unique element of $R$. This is easy to prove, and it yields

$End(R_R) \cong R$

so we get what we want: for any semisimple ring $R$ we have

$R \cong \bigoplus_{i = 1}^k M_{n_i}(D_i)$

for some division rings $D_i = End(X_i)$.

## Analysis

Most of what we did was show that the endomorphism ring of any object in a semisimple category is a finite product of matrix algebras over division rings. Only at the last minute did we use the fact that a semisimple ring $R$ actually is such an endomorphism ring, namely $End(R_R)$. So arguably the Wedderburn–Artin theorem is mainly about endomorphism rings of objects in semisimple categories. Let’s think about it that way.

Furthermore, while semisimple categories are necessarily abelian, we didn’t use much about abelian categories. We used some facts about kernels and cokernels, but only for proving Schur’s Lemma. And while the hom-sets of an abelian category are required to be abelian groups, we didn’t use subtraction anywhere.

So, let’s strip down the assumptions to see the essence of Wedderburn–Artin. For this I should review a bit of category theory — since I don’t really want nobody to understand me.

Let’s start with a semiadditive category: that is, a category with a zero object and biproducts. A zero object is an object that is both initial and terminal; we write it as $0$. Biproducts are what I’d been calling direct sums: the point is that they are both coproducts and products. More precisely, if two objects $X,Y$ in a category with a zero object have both a coproduct $X+Y$ and a product $X \times Y$, we get a canonical morphism $X + Y \to X \times Y$. If this is an isomorphism we call either $X + Y$ and $X \times Y$ the biproduct of $X$ and $Y$, and write it as $X \oplus Y$.

In a semiadditive category, for any objects $X$ and $Y$ the hom-set $Hom(X,Y)$ becomes a commutative monoid! In fact any semiadditive category is enriched over commutative monoids: for the proof go here or read this:

As a consequence, for any object $X$ in a semiadditive category, its set of endomorphisms $End(X)$ is not just a monoid but a rig. This is like a ring, but possibly without additive inverses. While mathematicians talk much more about division rings much more, the concept of a division rig makes perfect sense: it’s a rig where every nonzero element has a multiplicative inverse. For example, the nonnegative rational numbers, or the nonnegative elements of any ordered field, are a division rig.

We can also deal with matrices. Given a rig $R$ there’s a new rig $M_n(R)$ consisting of $n \times n$ matrices with entries in $R$, with matrix addition and multiplication defined in the usual way. Suppose $X$ is an object in a semiadditive category, and let $X^n$ stands for the $n$-fold biproduct $X \oplus \cdots \oplus X$. (This is both a coproduct and a product, so the notation $X^n$ is not so bad.) Then you can check that

$End(X^n) \cong M_n(X)$

This uses almost all the properties of biproducts.

Now that we’ve got division rigs and matrix rigs, we’re ready to state and prove an easy generalization of the Wedderburn–Artin theorem!

## Generalization

Suppose $\mathsf{A}$ is a semiadditive category. Say an object $X \in \mathsf{A}$ is simple if every nonzero $f \colon X \to X$ is an isomorphism. Say an object $X \in \mathsf{A}$ is semisimple if it can be written as a biproduct of simple objects $X_1, \dots, X_k$ with $hom(X_i,X_j) \cong \{0\}$. We allow each object $X_i$ to show up multiple times in this biproduct, so

$\displaystyle{ X \cong \bigoplus_{i = 1}^k X_i^{n_i} }$

for some positive integers $n_i$. These concepts of ‘simple’ and ‘semisimple’ reduce to the usual ones when $\mathsf{A}$ is abelian.

With these definitions, we get this result:

Abstract Wedderburn–Artin Theorem. If $R$ is a semisimple object in a semiadditive category $\mathsf{A}$, then $End(R)$ is a finite product of matrix rigs over division rigs.

Proof. Write

$\displaystyle{ R \cong \bigoplus_{i = 1}^k X_i^{n_i} }$

for simple objects $X_1, \dots, X_n$ with $hom(X_i,X_j) \cong \{0\}$, and then use the properties of biproducts to conclude

$\displaystyle{ End(R) \cong \bigoplus_{i = 1}^n M_{n_i}(End(X_i}))$

where each $End(X_i)$ is a division rig since $X_i$ is simple. ▮

When $\mathsf{A}$ is an abelian category, the division rings and the multiplicities $n_i$ are determined (up to permutation) by the object $X$. In this generalization I’m not claiming that, but I haven’t examined the issue yet.

I believe we can use this generalization to get a Wedderburn–Artin theorem for rigs. For this let $\mathsf{A} = Mod_R$ be the category of right modules of some rig $R$: that is, modules where it acts on commutative monoids. Take $R_R \in Mod_R$ to be $R$ as a right module over itself. Then I believe we have an isomorphism of rigs

$R \cong End(R_R)$

If so, when $R_R$ is a semisimple object in $Mod_R$ we can use this generalization of Wedderburn–Artin to write $R$ itself as a finite product of matrix rigs over division rigs.

I’m not claiming this generalization is a big deal. But taking apart a car engine and successfully putting it back together is the best way to get to know that engine — or so I’ve heard. And if you can put it back together with fewer pieces, and it still runs, that’s even better. So I felt I had to do this, to really understand Wedderburn–Artin.

Posted at June 14, 2023 4:34 PM UTC

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### Re: The Wedderburn–Artin Theorem

Yay for semiadditive categories.

Posted by: a on June 15, 2023 11:44 AM | Permalink | Reply to this

### Re: The Wedderburn–Artin Theorem

This is nice!

I think there are some typos in your statement and proof of Schur’s lemma. The second sentence of the statement (“Thus,…”) follows from the “It suffices to show” in the first sentence of the proof, but not (at least, not obviously) from the first sentence of the statement. And in the proof, it looks to me as though the roles of $0$ and $Y$ as cokernels are switched.

Posted by: Mike Shulman on June 16, 2023 6:50 AM | Permalink | Reply to this

### Re: The Wedderburn–Artin Theorem

You’re right on all counts. And the fixed version is even shorter!

Schur’s Lemma. If $X$ and $Y$ are simple objects in an abelian category, then any morphism $f \colon X \to Y$ is either an isomorphism or $0$. Thus, if $X$ is a simple object in an abelian category, $Hom(X,X)$ is a division ring.

Proof. If $X$ and $Y$ are simple the kernel of any morphism $f \colon X \to Y$ must be $0$ or $X$, and its cokernel must be $0$ or $Y$. If the kernel is $X$ or the cokernel is $Y$ we must have $f = 0$. The only alternative is that the kernel is $0$ and the cokernel is $0$. In an abelian category this implies that $f$ is an isomorphism. ▮

Posted by: John Baez on June 16, 2023 8:26 AM | Permalink | Reply to this

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