Past exam of the algebra course of the University of Oxford / 2009 / A / 3

A simple module is nonzero and has no proper nonzero submodule. A composition series of a module is a finite strict chain $0=V_0<V_1<\cdots <V_n=V$ of submodules with simple quotients $V_i/V_{i-1}$. These quotients are the composition factors of the series.

The Jordan–Hölder theorem for modules says that if a module has a composition series, then any two such series have the same length and the same simple factors up to isomorphism and reordering. In particular, every finite-dimensional module has a well-defined multiset of composition factors.

Choose $s\ne 0$ in the simple module $S$. The map $\varphi :A\to S$, $a\mapsto as$, is an $A$-module homomorphism whose image is the nonzero submodule $As$, hence all of $S$. Thus the first isomorphism theorem for modules givesThe kernel is a maximal left ideal because the quotient is simple. Whenever the regular module has finite length, as it does for a finite-dimensional algebra, take a composition series of the kernel and append $\ker \varphi <A$. Its last factor is $S$, so $S$ occurs as a composition factor of the regular module; the Jordan–Hölder theorem for modules makes that independent of the chosen series.

The quotient conclusion holds for every unital algebra. If an arbitrary infinite-dimensional regular module has no finite composition series, the universally valid statement is that $S$ is a simple subquotient; a claim about the factors of a finite series requires finite length.

A semisimple module is a direct sum of simple modules. An algebra is semisimple when all its modules are semisimple. Suppose the left regular module $A$ is semisimple. Every module $M$ is a quotient of a free module ${⨁}_{j\in J}A$, by sending its basis elements to a generating set of $M$. This free module is a direct sum of simple modules.

A quotient of a direct sum of simple modules is semisimple: the images of its simple summands are simple or zero and span the quotient. Choose a maximal direct sum among these simple images, allowing all simple submodules in the choice. If it does not span the whole quotient, some simple image is not contained in it. Its intersection with that direct sum is zero by simplicity, so it could be added, contradicting maximality. For infinite index sets, Zorn's lemma supplies the maximal direct sum; in finite dimension the construction terminates directly.

Hence every $M$ is semisimple, and $A$ is a semisimple algebra. This proves semisimplicity from the regular module.

The multiplication gives $v_1=1_A$ and $v_2^2+3v_2+2v_1=0$. Since $v_1,v_2$ form a basis,by the Chinese remainder theorem. Thus there are exactly two nonisomorphic simple modules, both one-dimensional. With basis vectors $s_{-1},s_{-2}$ their actions areThe two orthogonal primitive idempotents are $e_{-1}=2v_1+v_2$ and $e_{-2}=-v_1-v_2$. They sum to $v_1$ and span the two simple ideals. Consequently the regular module splits as $A=\mathbb{C}{e}_{-1}\oplus \mathbb{C}{e}_{-2}$, so $A$ is semisimple.

Its complete list of composition series isIndeed every proper nonzero submodule has dimension one and is an invariant line for multiplication by $v_2$. The matrix $\left(\begin{array}{cc}0& -2\\ 1& -3\end{array}\right)$ has the two distinct eigenvalues $-1,-2$, so these are exactly its two invariant lines.