Let $\lambda \in P^+$ and consider the Verma module $M(\lambda)$ with irreducible quotient $V(\lambda)$. So, there exists a maximal submodule $M\subseteq M(\lambda)$ such that $M(\lambda)/M \cong V(\lambda),$ whence we derive the exact sequence
$$0 \rightarrow M \rightarrow M(\lambda) \rightarrow V(\lambda)\rightarrow 0.$$
The above exact sequence does not split, since $M(\lambda)$ is indecomposable. Suppose even further that $M$ is a simple $\mathfrak g$-module. In this case, is $\mbox{Ext}_\mathfrak g (V(\lambda),M)$ 1-dimensional? That is, any non-splitting exact sequence $0\rightarrow M \rightarrow W \xrightarrow{\,\pi\,} V(\lambda) \rightarrow 0$ implies that $W\cong M(\lambda)$?
My efforts so far were the following: Let $u\in W$ be such that $\pi(u)$ is a highest weight vector of weight $\lambda$, so that $W' = U(\mathfrak g)u \subseteq W$ is a highest weight module of weight $\lambda$, and such that $V(\lambda)$ is isomorphic to a quotient of $W'$ by restriction of $\pi$ to $W'$. Since $M$ and $V(\lambda)$ are irreducible, from the exact sequence $0\rightarrow M \rightarrow W \rightarrow V(\lambda)$ we have that $W$ is a module of length $2$, with $W'$ being a module of lenght $\leq 2$ by consequence. If $W'$ is of length $1$, it is simple, whence it follows that $W'\cong V(\lambda)$, as $V(\lambda)$ is a quotient of $W'$. If $W'$ is of length $2$, then $W' = W$. The former case would imply that $0\rightarrow M \rightarrow W \xrightarrow{\,\pi\,} V(\lambda) \rightarrow 0 $ splits, so we must have $W= W'$. Therefore, $W'$ is a highest weight module of weight $\lambda$ and is, thus, isomorphic to a quotient of $M(\lambda)$. From the exact sequences $0\rightarrow M \rightarrow M(\lambda) \rightarrow V(\lambda)\rightarrow 0$ and $0 \rightarrow M \rightarrow W \rightarrow V(\lambda) \rightarrow 0$, we conclude that $W' = W$ and $M(\lambda)$ have the same dimension, so that $W'= W\cong M(\lambda)$.
Edit: The Verma Module is not finite dimensional. So I cannot state that $M(\lambda)$ and $W'=W$ have the same dimesion by the argument above...
Is this reasoning correct?