\hspace{1 mm}
      \bullet \hspace{1 mm} \bullet & \overset{g_{l-1}}{\bullet} \ar[dr]
      & & \overset{g_l}{\bullet} \ar[dl] \\
    &   & \underset{f_0}{\bullet} & & \underset{f_1}{\bullet} &  & &  & \underset{f_{l-1}}{\bullet} &
    }$$ Define a vector $V(\lambda): = g_0 + g_1 \lambda + g_2 \lambda^2 +
\cdots + g_l  \lambda^l$. A trivial verification shows that $(a +
\lambda b) (V(\lambda)) = 0$ and therefore $a + \lambda b$ is a singular transformation as desired.

Conversely, suppose $a + \lambda b$ is singular. Then there is a non-zero vector $V(\lambda) =  g_0 + g_1 \lambda + g_2 \lambda^2
\cdots g_l \lambda^l$ of smallest degree $l$ in $V_1 \otimes
k[\lambda]$ (so $g_l \ne 0$) such that $(a + b \lambda) (V(\lambda))
= 0$. This means: $a(g_0) = 0$, $b(g_i) = a(g_{i+1})$ for $0 \le i
\le l-1$, and $b(g_l) = 0$. We now argue that these equations give a copy of $\Omega^l(k)$ inside $M$. To this end, it suffices to show that the vectors $\{g_0, g_1, g_2, \cdots, g_l \}$ are linearly independent. As a further reduction, we claim that it suffices to show that $\{ a(g_1),  a(g_2), \cdots, a(g_l) \}$ are linearly independent. For, then it will be clear that $\{ g_1, g_2, \cdots,
g_l \}$is linearly 