An intersection of two primitive ideals need not be primitive

Given: $\mathfrak{g}={\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ with Casimir $\mathrm{\Omega }$, the finite-dimensional simple modules $L(0)=\mathbb{C}$ and $L(1)$ (the standard two-dimensional module), and the ideals $I_0={\mathrm{Ann}}_{U(\mathfrak{g})}L(0)$, $I_1={\mathrm{Ann}}_{U(\mathfrak{g})}L(1)$.

Annihilators of the simple modules $L(0)$ and $L(1)$ are primitive ideals (Annihilators of simple highest-weight modules are primitive, Primitive ideals of an enveloping algebra, The annihilator of a module over an enveloping algebra).

A primitive ideal $I$ satisfies $I\cap Z(U(\mathfrak{g}))=\ker {\chi }_I$, and this intersection is a maximal ideal of $Z(U(\mathfrak{g}))$ (A primitive ideal determines a central character, Prime ideals and maximal ideals in a commutative ring).

Put $\mathrm{\Omega }:=ef+fe+\frac{1}{2}{h}^2\in Z(U(\mathfrak{g}))$; it is central because the relations $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$ give $[h,\mathrm{\Omega }]=[e,\mathrm{\Omega }]=[f,\mathrm{\Omega }]=0$ (The special linear Lie algebra sl_2). The finite-dimensional simple modules are the $L(n)$, $n\ge 0$ (Finite-dimensional simple modules are classified by dominant highest weights), and on the highest-weight vector of $L(n)$ one has $ev=0$, $efv=[e,f]v+f(ev)=hv=nv$, $fev=0$, so $\mathrm{\Omega }v=(n+0+\frac{1}{2}{n}^2)v=\frac{n(n+2)}{2}v$; in particular ${\chi }_0(\mathrm{\Omega })=0$ and ${\chi }_1(\mathrm{\Omega })=\frac{3}{2}$ (Central character of a Lie algebra module).

In a commutative ring, two distinct maximal ideals have non-maximal intersection: if $M=\mathfrak{m}\cap \mathfrak{n}$ with $\mathfrak{m}\mathrm{\ne }\mathfrak{n}$ maximal were maximal, then $\mathfrak{m}\supseteq M$ and maximality of $M$ would force $M=\mathfrak{m}$, so $\mathfrak{m}\subseteq \mathfrak{n}$ and maximality of $\mathfrak{m}$ would force $\mathfrak{m}=\mathfrak{n}$, a contradiction (Prime ideals and maximal ideals in a commutative ring).