I had problems understanding the following proof. Maybe someone could help me with this?
Let {$\ x_i, y_i $} be the standard generators of the Lie Algebra L.
Let \ V'(\lambda)=U(L)/J'(\lambda).
Let \ V'(\lambda) be an irreducible standard cyclic module.
Let U(L) be the universal enveloping algebra.
Let \ J'(\lambda) be the left ideal generated by $\ I(\lambda)$ along with all $\ y_i^{m_i+1}.$ $\ I(\lambda)$ is the left ideal of U(L) generated by all $\ x_\alpha, (\alpha \succ 0)$, and by all $\ h_\alpha - \lambda(h_\alpha)1, (\alpha \in \Phi).$
Show that \ V'(\lambda) is finite dimensional.
For this it would suffice to show that it is a sum of finite dimensional $\ S_i$-submodules.
To show this we have to show that each $\ y_i $ is locally nilpotent on $\ V(\lambda) $.
This is obvious for the $\ x_i $, since we cannot have $\ \mu+k\alpha_i \prec \lambda$ for all $\ k \geq 0 $.
The coset of 1 in \ V'(\lambda) is killed by a suitable power of $\ y_i$ (namely, $\ m_i+1$).
\ V'(\lambda) is spanned by the cosets of all $\ y_{i_1}...y_{i_t}. (1\leq i_j \leq l) $.
If the coset of this monomial is killed by $\ y_i^k$, the coset of the longer monomial $\ y_{i_0}y_{i_1}...y_{i_t}$ is killed by $\ y_i^{k+3}$.
Induction on length of monomials, starting at 1, then proves the local nilpotence of $\ y_i$.
We have already proven before that $\ J(\lambda)$ is generated by $\ I(\lambda)$ along with all $\ y_i^{m+1}. m_i=\langle \lambda,\alpha_i \rangle, 1\leq i \leq l$, \lambda is a dominant integral linear function. For this we have assumed that \ V'(\lambda) is finite dimensional, which remains to show for the proof to be completed.
The whole proof is based upon the following theorem, which we have already proven before: If $\ \lambda \in H*$ is dominant integral, then the irreducible L-module $\ V=V(\lambda)$ is finite dimensional, and its set of weights $\ \Pi(\lambda)$ is permuted by the Weyl group, with $\ dimV_\mu=dimV_{\sigma\mu}$ for $\ \sigma \in$ Weyl group.