The Verma annihilator contains the central-character ideal

By [F1], $z\in Z(U(\mathfrak{g}))$ acts on $M(\lambda )$ by the scalar $\mathrm{pr}(z)(\lambda )={\chi }_{\lambda }(z)$. Hence $z\in \ker {\chi }_{\lambda }$ acts on $M(\lambda )$ by $0$, that is, $\ker {\chi }_{\lambda }\subseteq {\mathrm{Ann}}_{U(\mathfrak{g})}M(\lambda )$.

More generally, let $M=U(\mathfrak{g})v$ be any cyclic highest-weight module of highest weight $\lambda$ and let $u\in U(\mathfrak{g})$, $z\in \ker {\chi }_{\lambda }$. For $a\in U(\mathfrak{g})$ one has $(uz)(av)=u(zav)=u(azv)=ua{\chi }_{\lambda }(z)v=0$ by [F1] and centrality of $z$; since every element of $M$ has the form $av$, the element $uz$ annihilates $M$. As products $uz$ span $U(\mathfrak{g})\ker {\chi }_{\lambda }$, every element of that ideal annihilates every cyclic highest-weight module of highest weight $\lambda$.

Since ${\mathrm{Ann}}_{U(\mathfrak{g})}M(\lambda )$ is a two-sided ideal by [F3], it contains all products $uz$ with $u\in U(\mathfrak{g})$ and $z\in \ker {\chi }_{\lambda }$, hence contains the two-sided ideal $U(\mathfrak{g})\ker {\chi }_{\lambda }$ generated by $\ker {\chi }_{\lambda }$. Thus $U(\mathfrak{g})\ker {\chi }_{\lambda }\subseteq {\mathrm{Ann}}_{U(\mathfrak{g})}M(\lambda )$.

The Verma module $M(\lambda )$ has its unique simple quotient $L(\lambda )$ by [F2], and an operator annihilating $M(\lambda )$ annihilates each quotient by [F3], so ${\mathrm{Ann}}_{U(\mathfrak{g})}M(\lambda )\subseteq {\mathrm{Ann}}_{U(\mathfrak{g})}L(\lambda )=I(\lambda )$. Together with step 2.1 this gives the displayed chain, and step 1.2 gives the final assertion.