Past exam of the algebra course of the University of Oxford / 2011 / 2

The Jordan–Hölder theorem applies to a module of finite length: ifis a composition series of a module, its successive quotients are simple modules. Any two such composition series of a module have the same length $r$, and their composition factors are pairwise isomorphic after a permutation, with multiplicities preserved. Every finite-dimensional module over $A$ has finite length, since strict inclusions increase vector-space dimension. A finite-dimensional algebra over a field does not by itself make every possible module finite-dimensional, so the finite-length hypothesis should be retained in the statement.

For $a\in A$, $m\in M$,Additivity and complex linearity follow from those of $f$ and multiplication in $A$, so $f\cdot x$ is an module homomorphism. In particular the action stays inside $V$. Associativity givesBoth distributive laws hold pointwise. These are exactly the axioms for a right module, proving ${\mathrm{Hom}}_A(M,A)$ is a right $A$-module.

By the Artin–Wedderburn theorem, a finite-dimensional complex simple algebra is $M_d(\mathbb{C})$, with a unique simple module up to isomorphism, namely the column module $S={\mathbb{C}}^d$. Its left regular module is $S^{\oplus d}$. The Schur lemma gives ${\mathrm{End}}_A(S)=\mathbb{C}$, soThe right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have dimension $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne 0$, suitable matrix units send it to each coordinate row. Thus $V_M$ is a simple right $A$-module.

Write $e_{ij}$ for the matrix units. Left multiplication by $a=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ givesThus a composition series of a module isEach quotient is one-dimensional, hence a simple module. The required direct sum isThe first summand is simple. The second has a unique nonzero proper submodule, $\mathbb{C}{e}_{12}$: any vector $u{e}_{12}+v{e}_{22}$ with $v\ne 0$ generates $e_{12}$ by applying $e_{12}$ and then generates $e_{22}$. A decomposition into two nonzero summands would require a second one-dimensional submodule, which is impossible. Hence this summand is an indecomposable module.

Choose bases $S_1=\mathbb{C}{s}_1$, $S_2=\mathbb{C}{s}_2$ and setBoth are simple modules. They are nonisomorphic because $e_{11}$ acts as the identity on $S_1$ and as zero on $S_2$. In the ascending composition series of a module above, the factors are, in order,The last factor is represented by $e_{22}$ modulo the span of $e_{11},e_{12}$.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.