Past exam of the algebra course of the University of Oxford / 2008 / A

An algebra homomorphism is a $K$-linear map preserving multiplication, and preserving the unit under the unital convention. An algebra isomorphism is a bijective algebra homomorphism; its inverse then preserves the same operations.

A commutative domain algebra may map into a noncommutative target: take $A=K$, $B=K\times M_2(K)$, and $\phi (\lambda )=(\lambda ,\lambda I_2)$. This is even an injective unital homomorphism. The matrix factor makes the target noncommutative over every field, since $E_{12}{E}_{21}\ne E_{21}{E}_{12}$.

A noncommutative algebra may have a commutative quotient: take $A=K\times M_2(K)$, $B=K$, and $\phi (\lambda ,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.

For the matrix classifications, write $E_{ij}$ for a matrix unit and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results areIndeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x^{\prime },u^{\prime })=(x{x}^{\prime },x{u}^{\prime }+x^{\prime }u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+z{N}^2{)}^2=y^2{N}^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $K{E}_{12}$. Noncommutativity follows from $E_{22}{E}_{12}=0$ but $E_{12}{E}_{22}=E_{12}$.

The coordinate map $xI+y{E}_{12}+z{E}_{13}\mapsto xI+y{E}_{13}+z{E}_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only noncommutative member, and among the commutative members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an algebra isomorphism must carry one onto the other. Thus the only distinct isomorphic pair is $A_2\cong A_5$.

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

The character of a finite-dimensional complex representation is ${\chi }_{\rho }(g)=\mathrm{Tr}\rho (g)$; it is irreducible if the representation is irreducible. The character inner product isDistinct irreducible characters are orthogonal and each has norm one. For a permutation representation, the diagonal entry of the permutation matrix at a basis vector is one exactly when that vector is fixed. Its trace therefore counts fixed points.

Let $r$ be rotation through $\pi /2$. Let $s_x,s_y$ be reflections in the coordinate axes, and $s_d,s_{d^{\prime }}$ reflections in the two diagonals. The eight values areOnly diagonal reflections fix vertices, two each. Hence

Complete reducibility and character orthogonality imply that the squared multiplicities sum to three. The trivial constituent has multiplicity one, so the remaining two units come from two distinct nontrivial constituents, each with multiplicity one. Their degrees add to $4-1=3$, so they are two and one. ThusThe two-dimensional constituent is the complexification of the geometric action on the plane. Its trace is $2$ on the identity, $-2$ on the half-turn, zero on the quarter-turns, and zero on every reflection. Its character norm is $(4+4)/8=1$, proving it irreducible. Alternatively, the span of the two coordinate functions on the vertex set is a two-dimensional invariant subspace of the permutation module with this geometric action. This identifies the requested representation rather than merely its degree.