In an operator algebras class, the professor said the following.

Let $G$ be a finite group, and consider the complex group algebra it generates. Using the following 3 facts, we're supposed to see that $\mathbb{C}[G]$ is algebra-isomorphic to a direct sum of matrix algebras, each the image of some irreducible representation of $\mathbb{C}[G]$ I'm guessing by representation, he means an algebra homomorphism into the linear transforms on some vector space. By irreducible he probably means there are no invariant subspaces for the representation other than $0$ and the whole vector space.

3 facts are: 1. Artin Wedderburn 2. Schur's Lemma 3. Every division ring over $\mathbb{C}$ is trivial. (I assume what he means is every division algebra over $\mathbb{C}$ is algebra-isomorphic to $\mathbb{C}$.)

I am incredibly rusty at algebra, being an analyst. I have a number of questions. The first is if this kind of situation happens a lot in operator algebras. I was told when getting into operator algebras that the "algebra" part of the name of the field was illusory, and hence would not have to worry that my natural intuition and talents are in analysis, and not at all in algebra. Is it advisable to master the graduate level type of algebra? I have had no problem doing the algebra that has come up thus far. On the other hand, I've heard a professor of mine talking about how group algebras are an important tie-in to the field of operator algebras, so I wonder if the fact that I've forgotten nearly everything might merit a full review of the first year graduate type material in algebra.

Another question is if Artin Wedderburn really works for algebras. I know it's a statement for left/right semisimple rings (with unit), so one could hope that it would work for algebras with semisimple ring structure. Assuming it does I'm perplexed by the fact that matrix algebras are finite dimensional, so we could deduce that all simple algebras are finite dimensional. Or is this not the case because Artin Wedderburn only tells you that simple algebras are isomorphic to a matrix algebra over some division algebra over $\mathbb{C}$? In that case, how can I know that the matrix entries in this case are actually just complex?

Thirdly, how would I put together these 3 facts to see the claim?

Fourthly, why is every division algebra over $\mathbb{C}$ just a copy of $\mathbb{C}$?

  • $\begingroup$ In these notes, I have essentially quoted the professor, so on the occasions where I have expressed uncertainty in the correct translation of the statement, it is because I perceived that what he said is not the actual precise statement. Please correct me if any of my interpretations are wrong. Also, I realize some people may suggest regarding my first question that I review algebra as I need it, but I have a tough time deciding what I need, and then deciding what it's logically dependent on, what that's dependent on, etc. $\endgroup$
    – Jeff
    Sep 30, 2012 at 16:10
  • $\begingroup$ There are no division algebras over an algebraically closed field $K$ except $K$ itself. Suppose $D$ is such an algebra. Given $a\in D$, $a$ satisfies some polynomial equation of minimal degree with coefficients in $K$. Since $K$ is algebraically closed, it has a zero, say $r$. Factoring out $(x-r)$ and plugging in $a$ produces a contradiction if the degree of the polynomial is greater than 1, so $a\in K$. $\endgroup$
    – Potato
    Sep 30, 2012 at 16:53

2 Answers 2


First, let me point out that a matrix algebra $M_n(K)$ over a field $K$ has dimension $n^2<\infty$, and the classification of semisimple algebras provided by Artin-Wedderburn only cares about the finite-dimensional case (at least to my knowledge).

I confess that I don't see how to use Schur's Lemma in proving your claim. (Perhaps, Schur appears while proving A-W, indeed a useful tool in the proof is that a simple algebra is isomorphic to some $M_n(D)$; and to prove this, one uses Schur's Lemma somewhere. But, as I said, I don't know what your professor had in mind.) But let me argue as follows: one can show that every $K[G]$-module is semisimple if char $K$ does not divide $|G|$ (this is Maschke's Theorem). In particular, $K[G]$ is a semisimple algebra. Now use Artin-Wedderburn (which does hold, in particular, for algebras over a field!) to write \begin{equation} K[G]=\prod_{i=1}^tM_{n_i}(D_i) \end{equation} where $D_i$ are skew-fields over $K$. Now you can worry about your favorite situation: $K=\textbf C$. There are no skew-fields over $\textbf C$ because, in general, a skew-field over $K$ is necessarily contained in a maximal abelian extension (of finite degree) $L$ of $K$. But $\textbf C$ has no finite nontrivial extensions. Hence \begin{equation} \textbf C[G]=\prod_{i=1}^tM_{n_i}(\textbf C). \end{equation}

But what is $t$? What are the $n_i$'s? How to interpret the factors $M_{n_i}(\textbf C)$?

Well, in fact $t$ is the number of irreducible representations of $G$ (if it helps, it equals the number of conjugacy classes in $G$), and $n_i$ is the dimension of the $i$-th irreducible complex representation of $G$. So the usual formula $|G|=\sum_{i=1}^tn_i^2$ is reflected by the A-W decomposition: \begin{equation} |G|=\dim_\textbf C\textbf C[G]=\sum_{i=1}^t\dim_\textbf CM_{n_i}(\textbf C). \end{equation}

  • $\begingroup$ "[...]the classification of semisimple algebras provided by Artin-Wedderburn only cares about the finite-dimensional case". Indeed, note that the Artin-Wedderburn theorem concerns semisimple rings, i.e. Artinian rings with trivial Jacobson radical, and algebras are Artinian iff they are finite dimensional (at least assuming the existence of unity). $\endgroup$ Sep 30, 2012 at 18:26
  • $\begingroup$ Thank you @Miha Habič, I should have thought about that! $\endgroup$
    – Brenin
    Sep 30, 2012 at 18:29
  • $\begingroup$ As a side note, Wedderburn originally proved the classification theorem for finite dimensional algebras over a skew field, using the concept of a prime algebra and constructing a system of matrix units. Only later did Artin add the ring theoretic generalization. $\endgroup$ Sep 30, 2012 at 18:32

Every (finite-dimensional) division algebra over $\mathbb{C}$ is isomorphic to $\mathbb{C}$ because $\mathbb{C}$ is algebraically closed. We argue as follows. Let $A$ be a finite-dimensional division algebra over $\mathbb{C}$, and choose any nonzero $x \in A$. The set $\{1, x, x^2, x^3, \ldots\}$ is linearly dependent, which means there is a relation of the form $p(x) = 0$ where $p$ is in a polynomial in $\mathbb{C}$. Since $\mathbb{C}$ is algebraically closed, this factors into $(x-a_1)(x-a_2) \ldots (x-a_n) = 0$, where $a_i \in \mathbb{C}$. Since $A$ is a division algebra, this implies $x-a_i = 0$ for some $i$. Since $x$ was arbitrary, we conclude that $A$ is 1 dimensional, i.e., $A \cong\mathbb{C}$.

  • $\begingroup$ Dear Ted, could you explain how your answer settles with this one? $\endgroup$
    – Arrow
    Jun 26, 2018 at 21:35
  • $\begingroup$ A division algebra $A$ over $\mathbb{C}$ requires that the elements of $A$ commute with $\mathbb{C}$ (this is part of the definition of an algebra over a field), which is stronger than just saying that $A$ contains a subring isomorphic to $\mathbb{C}$ (as is assumed in the other question) . $\endgroup$
    – Ted
    Jun 27, 2018 at 4:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .