June 11, 2023

Brauer’s Lemma

Posted by John Baez

The Wedderburn–Artin Theorem has always been one of my friends in ring theory, not because I understand it well, but because it reduces a potentially scary general situation to one that fits pretty nicely inside my mathematical physics brain. One proof of it uses a cute result called ‘Brauer’s Lemma’.

The Wedderburn–Artin theorem says that if you have a semisimple ring $R$, it must be a finite product of matrix algebras over division rings. When my mathematical physicist brain sees this, it mutters: “oh, good — like $n \times n$ matrices with real, complex or quaternionic entries!” These are familiar from the foundations of quantum mechanics, the study of classical Lie groups, and so on.

But my pals $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$ are much more than division rings: they are division algebras over a field, namely the field of real numbers! Thus they pertain to a special case of the Artin–Wedderburn Theorem, which would have been familiar to Wedderburn. The full-fledged theorem breaks free from these shackles, and works with algebras over the integers: that is, rings. And lately, as I struggle to learn the rudiments of algebraic geometry, I’ve been trying to work over the integers myself.

But what’s a ‘semisimple’ ring, and how do they pull you into the study of division rings? And why is the title of this article ‘Brauer’s Lemma’?

Any ring $R$ has a category of (left) modules, say $R Mod$. A module is simple if it has no nontrivial submodules — or equivalently, no nontrivial quotient modules. Simple modules are like primes: they’re building blocks of more interesting modules. The easiest way to build more complicated modules is to take direct sums, so we say a module is semisimple if it’s a finite direct sum of simple modules.

The idea of ‘semisimplicity’ is one of those foundational notions that spreads through many areas of algebra. It’s a good name, because when you get used to algebra, the ‘semisimple’ objects tend to be the ones you can more easily understand. They may not be simple to work with, but at least they’re…. halfway simple.

So much for semisimple modules. What about rings?

A ring is semisimple if it’s semisimple as a left module over itself. And don’t worry: in case you’re thinking this should be called ‘left’ semisimple and there’s also a ‘right’ version, there’s a nice fact: a ring is semisimple as a left module over itself iff it’s semisimple as a right module over itself!

For example, think about $R = M_n(k)$, the ring of $n \times n$ matrices over some field. This has an obvious module, namely $k^n$. This module is simple, and every module is a direct sum of (possibly infinitely many) copies of this one. In particular, as a left module over itself, $M_n(k)$ is a direct sum of $n$ copies of $k^n$. That’s because when you multiply two $n \times n$ matrices $A$ and $B$, you’re really multiplying $A$ and a list of $n$ different column vectors, one for each column of $B$. So this ring is semisimple.

Puzzle 1. Show that $R = M_n(k)$ is semisimple if $k$ is any division ring: that is, a ring where every nonzero element has a multiplicative inverse.

Puzzle 2. Show that $R = M_n(k)$ is not semisimple if $k = \mathbb{Z}$.

So matrix algebras over division rings are semisimple. It’s also easy to see that a direct sum of two semisimple rings is semisimple. The Wedderburn–Artin Theorem says that’s all.

Wedderburn–Artin Theorem. Every semisimple ring is isomorphic to a finite direct sum of matrix algebras over division rings.

If you prefer (associative, unital) algebras over fields to rings, here’s another version:

Wedderburn–Artin Theorem for Algebras over Fields. Every semisimple algebra over a field $k$ is isomorphic to a finite direct sum of matrix algebras over division algebras over $k$.

I won’t bother to define ‘semisimple algebra’ and ‘division algebra’, except to say that they go just like the definitions of semisimple ring and division ring: just cross out ‘ring’ and write in ‘algebra over $k$’ everywhere.

Anyway, how do we prove Wedderburn–Artin? There are various proofs. I was naturally drawn to this one, due to the title:

A key step here is something called ‘Brauer’s Lemma’. I like it because it shows you one reason division rings get into the act.

Brauer’s Lemma. Suppose $R$ is a ring and $K \subseteq R$ is a minimal left ideal with $K^2 \ne 0$. Then $K = R e$ for some $e \in K$ with $e^2 = e$, and $e R e$ is a division ring.

Here a minimal left ideal is a nonzero left ideal for which the only smaller left ideal is $\{0\}$.

My mathematical physicist brain processes this as follows. Suppose $R$ is the ring of $n \times n$ matrices over some division ring like $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$. This has a minimal ideal $K$ consisting of matrices with just one nonzero column, say the $j$th column. You can see $K^2 \ne 0$. Then how do we write $K = R e$? We can take $e$ to be the matrix that’s zero everywhere except for a 1 in the $j$th row and $j$th column. Multiplying any matrix by this kills off everything except the $j$th column!

Clearly $e^2 = e$. Lastly, $e R e$ consists of matrices that are zero except in the $j$th row and $j$th column. So $e R e$ is isomorphic to the division ring we started with!

The reason I admire Brauer’s Lemma so much is that it works for any ring, and coughs up a division ring. Well, at least it works for any ring that has a minimal left ideal $K$ with $K^2 \ne 0$. There are certainly rings without minimal left ideals — but these are called nonartinian, a term of abuse. And there are plenty of rings with minimal left ideals $K$ that have $K^2 = 0$. For example, take $\mathbb{R}[x]/\langle x^2 \rangle$, and look at the ideal generated by $x$. But still, Brauer’s Lemma is ridiculously general.

How do you prove it? Luckily, no deep techniques or preliminary lemmas are required! You just need to follow your nose in a fairly skillful way, using the minimality condition a lot. I found Nicholson’s proof cryptically terse, so here’s a version with more steps filled in:

Proof of Brauer’s Lemma. Since $0 \ne K^2$, we must have $K u \ne 0$ for some $u \in K$. Of course $u \ne 0$. But $K u$ is a left ideal contained in $K$, so by minimality

$K u = K.$

Thus $u = e u$ for some $e \in K$. Now, let

$L = \{a \in K : a u = 0 \}$

$L$ is a left ideal since for any $r \in R$ we have

$a \in L \; \implies \; a u = 0 \; \implies \; r a u = 0 \; \implies \; r a \in L.$

Note $L \subseteq K$ by definition, so by the minimality of $K$ we must have either $L = 0$ or $L = K$. But $e$ is in $K$ but not in $L$, since $e u = u \ne 0$, so we must have $L = 0$. In other words

$a \in K \; and \; a u = 0 \quad \implies \quad a = 0.$

What can we do with this? Well, $u = e u$ so $e u = e^2 u$ so $(e - e^2)u = 0$, so let’s take $a$ above to be $e - e^2$. We conclude that $e - e^2 = 0$! So $e$ is what we call an idempotent:

$e^2 = e.$

Now we claim that $K = R e$. The reason is that $e$ is in the left ideal $K$, so $R e \subseteq K$. Since $K$ is minimal this implies either $R e = 0$ or $R e = K$. But $e \ne 0$ so $R e \ne 0$.

Finally, why is $e R e$ a division ring? Its unit is $e$, of course. Suppose $b \in e R e$ is nonzero. Why does it have an inverse? Well, $R b$ is a nonzero ideal contained in $R e = K$ so by minimality $R b = R e$. So, we must have $e = r b$ for some $r \in R$. But this implies $e r e$ is the left inverse of $b$ in $e R e$:

$(e r e) b = e r (e b) = e r b = e^2 = e.$

Finally, in a ring where every nonzero element has a left inverse, every element has a two-sided inverse, so it’s a division ring! To see this, say $b \ne 0$. It has a left inverse, which I’ll call $x$ now. Why is $x$ also the right inverse of $b$? Well, $x$ must also have its own left inverse, which I’ll call $y$. So we have $x b = 1$ and $y x = 1$, giving

$y = y (x b) = (y x) b = b .$

Thus $b$ is the left inverse of $x$. So $x$ is the right inverse of $b$. ▮

Whew! That’s algebra at its rawest. I used to hate this stuff. But it’s actually impressive how you can scrabble around with equations, using just addition and multiplication, and get something nontrivial.

If you think my proof was too long to understand, with too many small steps cluttering the overall logic, you might prefer Nicholson’s. Here’s how he says it:

Proof of Brauer’s Lemma. Since $0 \ne K^2$, certainly $K u \ne 0$ for some $u \in K$. Hence $K u = K$ by minimality, so $e u = u$ for some $e \in K$. If $r \in K$, this implies $r e - r \in L = \{a \in K \vert a u = 0\}$. Now $L$ is a left ideal, $L \subseteq K$, and $L \ne K$ because $e u \ne 0$. So $L = 0$ and it follows that $e^2 = e$ and $K = R e$.

Now let $0 \ne b \in e R e$. Then $0 \ne R b \subseteq R e$ so $R b = R e$ by minimality, say $e = r b$. Hence $(e r e)b = e r(e b) = e r b = e^2 = e$, so $b$ has a left inverse in $e R e$. It follows that $e R e$ is a division ring. ▮

Posted at June 11, 2023 9:47 PM UTC

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Re: Brauer’s Lemma

a ring is semisimple as a left module over itself iff it’s semisimple as a left module over itself!

One of those lefts should be right?

Posted by: Elias Gabriel Amaral da Silva on June 12, 2023 12:37 AM | Permalink | Reply to this

Re: Brauer’s Lemma

Yes! Although the statement is true as it stands.

Thanks, I’ll fix it.

Posted by: John Baez on June 12, 2023 3:59 PM | Permalink | Reply to this

Re: Brauer’s Lemma

I always preferred the module-theoretic proof, using Schur’s Lemma: If $S$ and $T$ are simple modules, then every homomorphism $S\to T$ is either zero or an isomorphism.

From this we get that $\mathrm{End}_R(S)$ is a division algebra, and if $X$ is a finite direct sum of simple modules, then $\mathrm{End}_R(X)$ is a finite product of matrices over division rings.

More precisely, write $X=S_1^{d_1}\oplus\cdots\oplus S_r^{d_r}$ with the $S_i$ simple and pairwise nonisomorphic. Set $D_i=\mathrm{End}_R(S_i)$. Then $\mathrm{End}_R(X)=M_{d_1}(D_1)\times\cdots\times M_{d_r}(D_r)$.

In particular we see that the division rings which occur are the endomorphism rings of the simples which occur, and their multiplicities give the sizes of the matrix rings.

Finally, for any ring $R$ we have $R=\mathrm{End}_R(R)$, using the map sending $r$ to left multiplication by $r$. (I should point out here that I am using right $R$-modules and writing endomorphisms on the left, to do away with any pesky opposite signs.)

It follows that if $R$ is semisimple as a module, then $R$ is a finite product of matrix rings over division rings (coming from the simple direct summands and their multiplicities).

Posted by: Andrew Hubery on June 12, 2023 7:00 AM | Permalink | Reply to this

Re: Brauer’s Lemma

Thanks! Now I actually understand the Wedderburn–Artin theorem.

This is infinitely better than Nicholson’s proof, which I ran into when looking for a good proof. It’s also cleaner-looking than another proof I’d run into: Knap’s proof in Wedderburn–Artin ring theory, a chapter from his book Advanced Algebra. But now that I get the point, I see Knap is doing essentially the same thing. It looks cluttered with unnecessary detail, but Nicholson would have said the same about my version of his proof of Brauer’s Lemma! The extra detail may be good for some students.

The thing that initially repelled me about Knap’s proof is that he uses the Jordan–Hölder theorem, which seems a bit “heavy”. But he uses it to go a bit further than your statement: he’s proving the uniqueness of the decomposition of a semisimple ring $R$ into simple left $R$-modules, and thus the uniqueness of its expression in terms of matrix algebras over division rings.

Now that I’m thinking about it more clearly, it seems Jordan–Hölder should become a lot easier in this semisimple case. But I suppose if you’re writing a textbook and you’ve already proved Jordan–Hölder, you’ll just use it.

(I’m trying to generalize Wedderburn–Artin, so I want to understand the logical prerequisites.)

In short: Brauer’s Lemma now seems largely irrelevant to the best proof(s) of Wedderburn–Artin. And yet it’s interesting. In your favorite proof, division rings show up as the endomorphism rings of simple objects in the category of $R$-modules, thanks to Schur’s Lemma. Concretely these simple objects are minimal one-sided ideals of $R$. Brauer’s Lemma emphasizes that these division rings also show up as subrings $e R e \subseteq R$ for idempotents $e$. But this extra fact is also easy to get, once one notes that these minimal one-sided ideals must be of the form $e R$ (or $R e$, depending on which side you like).

Posted by: John Baez on June 12, 2023 4:34 PM | Permalink | Reply to this

Re: Brauer’s Lemma

While I have your ear, Andrew: the proof of Wedderburn–Artin on Wikipedia claims the division rings $D_i$ are not determined by the semisimple ring $R$. Aren’t they? (Only up to reordering, of course.)

I hadn’t seen this proof at first, since it’s in the article “Decomposition of a module”. Their article “Wedderburn–Artin theorem” contained a different proof — which was completely incorrect! Various editors had complained about it, but nobody had gotten around to removing it. So I removed it, and now I’m polishing up all their material on the Wedderburn–Artin theorem.

Posted by: John Baez on June 12, 2023 5:00 PM | Permalink | Reply to this

Re: Brauer’s Lemma

Lam’s A first course in noncommutative rings (2001: p. 33–34) also proves it like this, and shows that in a decomposition $R \cong \operatorname{M}_{n_1}(D_1)\times \dots \times \operatorname{M}_{n_r}(D_r)$ the division rings $D_i$ and multiplicities $r_i$ are uniquely determined.

Posted by: AT on June 12, 2023 8:27 PM | Permalink | Reply to this

Re: Brauer’s Lemma

I’ve only just read part of Knap’s chapter (the second edition is on Project Euclid, BTW) and actually the statement (including the uniqueness of the $D_i$’s and $n_i$’s) and proof is the same as Lam’s, Jordan–Hölder and all… 😅

Posted by: AT on June 12, 2023 8:42 PM | Permalink | Reply to this

Re: Brauer’s Lemma

An easy way to get uniqueness is as follows. Take any simple module $S$, with endomorphism ring $D$, a division algebra. Then $\mathrm{Hom}_R(R,S)$ is a non-zero right $D$-module, and the multiplicity of $S$ as a direct summand of $R$ is the $D$-dimension of the hom space, so is intrinsic.

Posted by: Andrew Hubery on June 12, 2023 11:05 PM | Permalink | Reply to this

Re: Brauer’s Lemma

Okay, thanks folks! I fixed the accursed Wikipedia entry that claimed that a semisimple ring does not uniquely specify the division rings from which it’s built. The articles on simple rings, semisimple rings and the Wedderburn–Artin theorem seemed to contain more bullshit than most of the math articles I’ve read on Wikipedia. But maybe it’s just because I’ve been intensively editing these articles and thus noticing the mistakes. I suppose I should read everything there with more skepticism.

I believe I’ve fixed most of the mistakes and helped clarify the horrible fact: not every simple ring is semisimple.

Posted by: John Baez on June 13, 2023 7:33 PM | Permalink | Reply to this

Knapp

Missing the extra p’s in that name. (No relation, I think.)

Re: Brauer’s Lemma

Off topic, but I’m going to relay something I recently learned from my colleague Harry Braden: the first letter of Wedderburn’s name is pronounced W, not V.

Like me, Wedderburn has a name that people assume is Germanic but isn’t. Joseph Wedderburn was Scottish. Lots of place names in Scotland, Ireland and northern England have a “burn” in them (e.g. Blackburn), and a “burn” itself is a stream. I’d never come across the word “wedder”, but here’s the dictionary entry.

Posted by: Tom Leinster on June 12, 2023 5:33 PM | Permalink | Reply to this

Re: Brauer’s Lemma

Offer topic: the Springer resolution is not pronounced the same way as Springer-Verlag.

Posted by: Allen Knutson on June 13, 2023 3:14 AM | Permalink | Reply to this

Re: Brauer’s Lemma

Off topic, but I’m going to relay something I recently learned from my colleague Harry Braden: the first letter of Wedderburn’s name is pronounced W, not V.

Oh no! It’s going to take a long time to undo this habit!

a “burn” itself is a stream.

Okay, this’ll be my mnemonic: some burns are drier, while others are wedder.

Posted by: John Baez on June 13, 2023 7:07 AM | Permalink | Reply to this

Re: Brauer’s Lemma

Wedderburn has a name that people assume is Germanic

Really? When I see that name I can’t imagine it coming from anywhere but the British isles. (Neither ‘dd’ nor ‘urn’ is a common letter combination in German.) But I also have never heard anyone say the name out loud to my recollection, so I wasn’t misled by anyone else’s misconstrual.

I have rather the opposite problem, personally. My last name is German, and pronounced accordingly. But most Anglophones don’t recognize it as such, and make up all kinds of different pronunciations, many of which don’t obey the usual phonetics of any language.

Posted by: Mark Meckes on June 13, 2023 8:17 AM | Permalink | Reply to this

Re: Brauer’s Lemma

Maybe I’m projecting, though I’ve certainly heard other people pronounce it “Vedderburn”. As someone with only a little knowledge (and you know what they say about that), I thought it seemed plausibly German. Wieder is a German word, right? And ur is an acceptable letter combination (Urlaub). But now I know better!

Posted by: Tom Leinster on June 13, 2023 11:56 AM | Permalink | Reply to this

Re: Brauer’s Lemma

I probably just don’t remember not knowing German enough to have a strong instictive feeling about what looks plausibly German (though I’m far from fluent these days). Wederbern would look like a German name to me. So would Widerburg or many other variations. But Wedderburn, nope.

Of course I’d be just as susceptible as anyone else to making bad assumptions of the same type where languages I don’t know are concerned.

Posted by: Mark Meckes on June 13, 2023 5:33 PM | Permalink | Reply to this

Re: Brauer’s Lemma

I know German but I always say “Vedderburn”, even though I should know better. Especially since somebody told me he studied at the University of Edinburgh!

Posted by: John Baez on June 13, 2023 7:17 PM | Permalink | Reply to this

Re: Brauer’s Lemma

Fwiw, the definition of division ring in Puzzle 1 is wrong.

Posted by: Graham on June 13, 2023 2:03 PM | Permalink | Reply to this

Re: Brauer’s Lemma

Whoops — I don’t know what came over me! Thanks, I’ll fix that.

A ring with

$a b = 0 \; \implies \; a = 0 \; or \; b = 0$

is called a (possibly noncommutative) domain. So, if you like tricky prose, you can define a domain to be a ring with no nonzero zero divisors.

It’s definitely not true that matrices with entries in a domain form a semisimple ring. Consider 1×1 matrices with entries in $\mathbb{Z}$.

Posted by: John Baez on June 13, 2023 7:12 PM | Permalink | Reply to this

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