= Solution
The multiplication gives $v_1=1_A$ and $v_2^2+3v_2+2v_1=0$. Since $v_1,v_2$ form a <basis>,
$$
A\cong\mathbb C[t]/((t+1)(t+2))\cong\mathbb C\times\mathbb C
$$
by the <Chinese remainder theorem>. Thus there are exactly \b[two nonisomorphic simple <modules>], both one-dimensional. With basis vectors $s_{-1},s_{-2}$ their actions are
$$
\boxed{v_1s_{-1}=s_{-1},\quad v_2s_{-1}=-s_{-1};\qquad v_1s_{-2}=s_{-2},\quad v_2s_{-2}=-2s_{-2}.}
$$
The 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 $A=\mathbb Ce_{-1}\oplus\mathbb Ce_{-2}$, so \b[$A$ is semisimple].
Its complete list of <composition series> is
$$
\boxed{0<\mathbb C(2v_1+v_2)<A,\qquad0<\mathbb C(v_1+v_2)<A.}
$$
Indeed every proper nonzero submodule has dimension one and is an invariant line for multiplication by $v_2$. The matrix $\begin{pmatrix}0&-2\\1&-3\end{pmatrix}$ has the two distinct eigenvalues $-1,-2$, so these are exactly its two invariant lines.
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