Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 5 / 1

We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.

The Artin–Wedderburn theorem here says that there are positive integers $d_1,\dots ,d_r$ withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions $d_i$, and the factor sizes are unique up to reordering.

Here is a proof. Decompose the regular left module as ${}_{A}A={⨁}_{i=1}^r{S}_i^{\oplus n_i}$ with pairwise nonisomorphic simple modules $S_i$. These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$ and ${\mathrm{End}}_A(S_i)=\mathbb{C}$. For the latter assertion, an endomorphism has an eigenvalue over $\mathbb{C}$, and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at $1$, so ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{op}}$. Taking opposite algebras and using matrix transposition gives the asserted decomposition with $d_i=n_i$.

For a matrix algebra $M_d(\mathbb{C})$, the matrix units show that its only simple module is ${\mathbb{C}}^d$: the spaces $E_{jj}S$ are isomorphic through $E_{ij}$, and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of $S$. In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.

Assume first that $S\ne 0$, as required for the irreducible-module conclusion. If $B\to {\mathrm{End}}_{\mathbb{C}}(S)$ is surjective, any nonzero invariant subspace contains every image of one of its nonzero vectors under all endomorphisms, hence is all of $S$. Thus $S$ is irreducible.

Conversely an irreducible $B$-module is finite dimensional: for $s\ne 0$, $S=Bs$ is a quotient of the finite-dimensional vector space $B$. Let $D$ be the image of $B$ in ${\mathrm{End}}_{\mathbb{C}}(S)$. It acts faithfully and irreducibly. The subspace $J(D)S$ is a submodule, so is zero or all of $S$. The latter alternative would imply $J(D{)}^{k}S=S$ for every $k$, contradicting nilpotence of the Jacobson radical. Therefore $J(D)S=0$, and faithfulness gives $J(D)=0$.

The Artin–Wedderburn theorem makes $D$ a product of matrix algebras. A faithful simple module forces there to be just one factor, because all other factors would annihilate that module. Thus $D\cong M_d(\mathbb{C})$ and $S\cong {\mathbb{C}}^d$, so the action is the full endomorphism algebra. This is the Burnside matrix-algebra theorem.

For nonzero $S$, surjectivity is equivalent to irreducibility. The zero module is a literal exception if it is admitted: its endomorphism algebra is zero, so the action map is surjective, whereas the zero module is not irreducible.

Maschke's theorem makes the group algebra $\mathbb{C}G$ semisimple, since $G$ is a finite group and the ground field has characteristic zero. By the Artin–Wedderburn theorem, write $\mathbb{C}G\cong {\prod }_{i=1}^r{M}_{d_i}(\mathbb{C})$. Its simple modules give exactly the $r$ nonisomorphic irreducible complex group representations.

The center of each matrix algebra consists of scalar matrices, so ${\dim }_{\mathbb{C}}Z(\mathbb{C}G)=r$. On the other hand, an element ${\sum }_g{a}_{g}g$ is central precisely when its coefficients are constant on conjugacy classes. The sums of the elements in the separate conjugacy classes are therefore a basis of the center. Hence the number of irreducible complex representations equals the number of conjugacy classes.