The dominant cyclic generator survives

Let $J\subseteq U(\mathfrak{g})$ be the left ideal generated by ${\mathfrak{n}}^{+}$ and the elements $H-\lambda (H)$ with $H\in \mathfrak{h}$, and let $N=U(\mathfrak{g})\mathrm{/}J$ with $v_0=1+J$; then ${\mathfrak{n}}^{+}\cdot v_0=0$ and $H\cdot v_0=\lambda (H)v_0$ for $H\in \mathfrak{h}$, and by [L1] every element of $N$ is a linear combination of vectors $u\cdot v_0$ with $u\in U({\mathfrak{n}}^{-})$, so the linear map $\phi :U({\mathfrak{n}}^{-})\to N$, $\phi (u)=u\cdot v_0$, is onto.

The map $\phi$ is also injective. Let $\mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$. The linear functional ${\chi }_{\lambda }:\mathfrak{b}\to \mathbb{C}$ defined by ${\chi }_{\lambda }(H+x)=\lambda (H)$ is a Lie-algebra homomorphism to the abelian Lie algebra $\mathbb{C}$, because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Its multiplicative extension to the tensor algebra kills every relation $XY-YX-[X,Y]$, so the quotient definition in [L2] makes it an algebra homomorphism ${\chi }_{\lambda }:U(\mathfrak{b})\to \mathbb{C}$. By the PBW basis of [L1] every $u\in U(\mathfrak{g})$ has a unique finite expansion $u={\sum }_A{f}^A{u}_A$, with $u_A\in U(\mathfrak{b})$ and $f^A$ ranging over the monomials in a basis of ${\mathfrak{n}}^{-}$. Define $\rho (u)={\sum }_A{\chi }_{\lambda }(u_A)f^A$. Then $\rho$ is the identity on $U({\mathfrak{n}}^{-})$ and vanishes on $J$: for $x\in {\mathfrak{n}}^{+}$ and $H\in \mathfrak{h}$, one has ${\chi }_{\lambda }(x)=0$ and ${\chi }_{\lambda }(H-\lambda (H)1)=0$, so $\rho (ux)=\rho (u(H-\lambda (H)1))=0$ for every $u\in U(\mathfrak{g})$. Thus $U({\mathfrak{n}}^{-})\cap J=0$, and $\phi$ is a linear isomorphism by step 1.1. In particular $v_0\mathrm{\ne }0$, and the action of $U({\mathfrak{n}}^{-})$ on $N$ corresponds under $\phi$ to left multiplication.

For each $i$ set $w_i:=f_i^{m_i+1}\cdot v_0=\phi (f_i^{m_i+1})$; this is nonzero by [L6] and step 2.1, and it has weight $\lambda -(m_i+1){\alpha }_i$.

For every $i$ the vector $w_i$ satisfies ${\mathfrak{n}}^{+}\cdot w_i=0$: for $j=i$ one has $e_i{w}_i=[e_i,f_i^{m_i+1}]\cdot v_0$ by the product rule [L2] and $e_i\cdot v_0=0$, and the ${\mathfrak{s}\mathfrak{l}}_2$-commutation identity $[e_i,f_i^n]=n{f}_i^{n-1}(h_{{\alpha }_i}-(n-1))$ gives $[e_i,f_i^{m_i+1}]v_0=(m_i+1)f_i^{m_i}(h_{{\alpha }_i}-m_i)\cdot v_0=0$ because $h_{{\alpha }_i}\cdot v_0=m_i{v}_0$; for $j\mathrm{\ne }i$ one has $[e_j,f_i]=0$ by [L5], hence $[e_j,f_i^{m_i+1}]=0$ and $e_j{w}_i=f_i^{m_i+1}\cdot (e_j\cdot v_0)+[e_j,f_i^{m_i+1}]\cdot v_0=0$. Thus every simple generator $e_j$ kills $w_i$. The action is a Lie-algebra homomorphism, so a bracket of operators that each kill $w_i$ also kills $w_i$; since the $e_j$ generate ${\mathfrak{n}}^{+}$ by [L5], every element of ${\mathfrak{n}}^{+}$ kills $w_i$.

By steps 3.1 and 4.1 each $w_i$ is a highest weight vector of weight $\lambda -(m_i+1){\alpha }_i$, so by [L4] its submodule $K_i=U(\mathfrak{g})w_i$ has all weights $\le \lambda -(m_i+1){\alpha }_i$; since $(m_i+1){\alpha }_i$ is a nonzero element of $Q_{+}$ by [L6] and $\lambda -(\lambda -(m_i+1){\alpha }_i)=(m_i+1){\alpha }_i\mathrm{\notin }-Q_{+}$, no weight of $K_i$ equals $\lambda$, and $w_i\mathrm{\ne }0$.

Let $K={\sum }_i{K}_i$. A vector of weight $\lambda$ in a sum of submodules lies in the sum of their $\lambda$-weight spaces, each of which is zero by step 5.1, so $K$ has no weight $\lambda$; in particular $v_0\mathrm{\notin }K$.

The left ideal $I_{\lambda }$ defining $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )$ equals $J+{\sum }_{i}U(\mathfrak{g})f_i^{m_i+1}$, because it is the left ideal generated by the generators of $J$ together with the elements $f_i^{m_i+1}$; passing to the quotient by $J$ and using the isomorphism of step 2.1 gives $I_{\lambda }\mathrm{/}J={\sum }_i(U(\mathfrak{g})f_i^{m_i+1}+J)\mathrm{/}J={\sum }_i{K}_i=K$.

Therefore $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )=U(\mathfrak{g})\mathrm{/}{I}_{\lambda }=N\mathrm{/}K$, and the canonical generator $v_{\lambda }=v_0+K$ is nonzero by step 6.1.

Finally $v_{\lambda }$ has weight $\lambda$, because $H\cdot v_{\lambda }=\lambda (H)v_{\lambda }$ holds in $N$ and passes to the quotient, and ${\mathfrak{n}}^{+}\cdot v_{\lambda }=0$ for the same reason; hence $v_{\lambda }$ is a nonzero weight vector of weight $\lambda$ killed by ${\mathfrak{n}}^{+}$.

The class of the unit in $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )$ is nonzero, has weight $\lambda$, and is killed by ${\mathfrak{n}}^{+}$, as asserted.