[proofplan] The key point is that every proper submodule of a Verma module misses the highest weight line $\mathbb C v_\lambda$, because $v_\lambda$ cyclically generates $M(\lambda)$. We then use the weight-space decomposition of $M(\lambda)$ to prove that the sum of all proper submodules still misses this highest weight line, so that this sum is itself proper. This sum is therefore the unique maximal proper submodule, and quotienting by it gives the desired irreducible highest weight quotient. [/proofplan] [step:Record the highest weight line and cyclic generation of $M(\lambda)$] Let $\mathbb C_\lambda$ denote the one-dimensional $\mathfrak b$-module on which every $h\in\mathfrak h$ acts by the scalar $\lambda(h)$ and every $x\in\mathfrak n^+$ acts by $0$. By definition, \begin{align*} M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda, \end{align*} and its canonical vector is \begin{align*} v_\lambda:=1\otimes 1\in M(\lambda). \end{align*} For every $h\in\mathfrak h$ and $x\in\mathfrak n^+$, \begin{align*} h v_\lambda=\lambda(h)v_\lambda \end{align*} and \begin{align*} x v_\lambda=0. \end{align*} Moreover $M(\lambda)=U(\mathfrak g)v_\lambda$ by the definition of the induced module. Hence any $\mathfrak g$-submodule $N\subset M(\lambda)$ with $v_\lambda\in N$ satisfies \begin{align*} M(\lambda)=U(\mathfrak g)v_\lambda\subset N, \end{align*} so $N=M(\lambda)$. [/step] [step:Show every proper submodule has zero $\lambda$-weight space] Let $M(\lambda)_\mu$ denote the $\mu$-weight space of $M(\lambda)$: \begin{align*} M(\lambda)_\mu:=\{m\in M(\lambda): hm=\mu(h)m\text{ for every }h\in\mathfrak h\}. \end{align*} Since $\mathfrak g$ is finite-dimensional complex semisimple with the chosen positive root system $\Phi^+$ and triangular decomposition from the theorem statement, the hypotheses of [citetheorem:9376] apply to $M(\lambda)$. Therefore the Verma module has weight decomposition \begin{align*} M(\lambda)=\bigoplus_{\beta\in Q_+}M(\lambda)_{\lambda-\beta}, \end{align*} and its highest weight space is \begin{align*} M(\lambda)_\lambda=\mathbb C v_\lambda. \end{align*} Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule. We first note that $N$ is a weight submodule. Indeed, take $m\in N$ and write its finite weight decomposition as \begin{align*} m=\sum_{\mu\in S}m_\mu, \end{align*} where $S\subset\mathfrak h^*$ is a finite set and $m_\mu\in M(\lambda)_\mu$. Fix $\mu_0\in S$. For every $\mu\in S\setminus\{\mu_0\}$ choose $h_\mu\in\mathfrak h$ such that $\mu_0(h_\mu)\ne\mu(h_\mu)$, and define an element $p_{\mu_0}\in U(\mathfrak h)$ by \begin{align*} p_{\mu_0}:=\prod_{\mu\in S\setminus\{\mu_0\}}\frac{h_\mu-\mu(h_\mu)}{\mu_0(h_\mu)-\mu(h_\mu)}. \end{align*} Since $N$ is stable under $U(\mathfrak g)$ and $U(\mathfrak h)\subset U(\mathfrak g)$, we have $p_{\mu_0}m\in N$. On each weight component, $p_{\mu_0}$ acts by scalar evaluation at the corresponding weight, so $p_{\mu_0}m=m_{\mu_0}$. Thus each $m_{\mu_0}$ lies in $N$, and \begin{align*} N=\bigoplus_{\mu\in\mathfrak h^*}(N\cap M(\lambda)_\mu). \end{align*} Now suppose $N$ is proper. If $N\cap M(\lambda)_\lambda\ne 0$, then $M(\lambda)_\lambda=\mathbb C v_\lambda$ implies $v_\lambda\in N$, and the previous step gives $N=M(\lambda)$, a contradiction. Therefore \begin{align*} N\cap M(\lambda)_\lambda=0. \end{align*} [guided] The delicate point is to justify that a submodule cannot hide a highest-weight component inside a sum of lower-weight terms. This is why we first prove that submodules are compatible with the weight decomposition. Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule, and let $m\in N$. We use [citetheorem:9376], whose hypotheses are satisfied because the theorem statement assumes that $\mathfrak g$ is finite-dimensional complex semisimple, that $\mathfrak h$ is a Cartan subalgebra, and that a positive root system $\Phi^+$ has been fixed. It gives the direct weight-space decomposition \begin{align*} M(\lambda)=\bigoplus_{\beta\in Q_+}M(\lambda)_{\lambda-\beta}. \end{align*} Thus the vector $m$ has a finite decomposition \begin{align*} m=\sum_{\mu\in S}m_\mu, \end{align*} where $S\subset\mathfrak h^*$ is finite and $m_\mu\in M(\lambda)_\mu$. We want to prove that each component $m_\mu$ is itself in $N$. Fix one weight $\mu_0\in S$. Distinct linear functionals on $\mathfrak h$ can be separated by some element of $\mathfrak h$, so for each $\mu\in S\setminus\{\mu_0\}$ choose $h_\mu\in\mathfrak h$ satisfying \begin{align*} \mu_0(h_\mu)\ne\mu(h_\mu). \end{align*} Define \begin{align*} p_{\mu_0}:=\prod_{\mu\in S\setminus\{\mu_0\}}\frac{h_\mu-\mu(h_\mu)}{\mu_0(h_\mu)-\mu(h_\mu)}\in U(\mathfrak h). \end{align*} This element acts on a vector of weight $\nu$ by the scalar obtained by evaluating the displayed polynomial at $\nu$. Hence it acts as the identity on the $\mu_0$-weight component and as zero on each $\mu$-weight component with $\mu\in S\setminus\{\mu_0\}$. Therefore \begin{align*} p_{\mu_0}m=m_{\mu_0}. \end{align*} Because $N$ is a $\mathfrak g$-submodule, it is stable under the action of $U(\mathfrak g)$; since $p_{\mu_0}\in U(\mathfrak h)\subset U(\mathfrak g)$, the vector $p_{\mu_0}m$ belongs to $N$. Thus $m_{\mu_0}\in N$. Since $\mu_0$ was arbitrary, every weight component of every vector of $N$ belongs to $N$, and so \begin{align*} N=\bigoplus_{\mu\in\mathfrak h^*}(N\cap M(\lambda)_\mu). \end{align*} Now assume $N$ is proper. The same application of [citetheorem:9376] gives the one-dimensional highest weight space \begin{align*} M(\lambda)_\lambda=\mathbb C v_\lambda. \end{align*} If $N\cap M(\lambda)_\lambda$ were nonzero, then this one-dimensional space would force $v_\lambda\in N$. Since $v_\lambda$ cyclically generates $M(\lambda)$, this would imply \begin{align*} M(\lambda)=U(\mathfrak g)v_\lambda\subset N, \end{align*} so $N=M(\lambda)$, contradicting properness. Therefore every proper submodule has zero $\lambda$-weight space: \begin{align*} N\cap M(\lambda)_\lambda=0. \end{align*} [/guided] [/step] [step:Form the sum of all proper submodules and prove it is proper] Let $\mathcal S$ be the set of all proper $\mathfrak g$-submodules of $M(\lambda)$, and define \begin{align*} J(\lambda):=\sum_{N\in\mathcal S}N\subset M(\lambda). \end{align*} This is a $\mathfrak g$-submodule because sums of $\mathfrak g$-submodules are $\mathfrak g$-submodules. We claim that $J(\lambda)$ is proper. Since every $N\in\mathcal S$ is a weight submodule and satisfies $N\cap M(\lambda)_\lambda=0$, the $\lambda$-weight space of the sum is \begin{align*} J(\lambda)\cap M(\lambda)_\lambda=\sum_{N\in\mathcal S}(N\cap M(\lambda)_\lambda)=0. \end{align*} In particular $v_\lambda\notin J(\lambda)$. Since $v_\lambda\in M(\lambda)$, this proves \begin{align*} J(\lambda)\ne M(\lambda). \end{align*} Thus $J(\lambda)$ is a proper submodule of $M(\lambda)$. [/step] [step:Identify $J(\lambda)$ as the unique maximal proper submodule] By construction, every proper submodule $N\subset M(\lambda)$ is contained in $J(\lambda)$. The previous step shows that $J(\lambda)$ is itself proper. Therefore $J(\lambda)$ is a maximal proper submodule: if $J(\lambda)\subset K\subset M(\lambda)$ and $K$ is a proper submodule, then $K\in\mathcal S$, so $K\subset J(\lambda)$, hence $K=J(\lambda)$. It is unique with this property. If $J'$ is any maximal proper submodule of $M(\lambda)$, then $J'\in\mathcal S$, so $J'\subset J(\lambda)$. Since $J(\lambda)$ is proper and $J'$ is maximal proper, this inclusion forces \begin{align*} J'=J(\lambda). \end{align*} [/step] [step:Quotient by $J(\lambda)$ and verify irreducibility] Define the quotient module \begin{align*} L(\lambda):=M(\lambda)/J(\lambda), \end{align*} and let $\pi:M(\lambda)\to L(\lambda)$ be the quotient homomorphism. Since $J(\lambda)$ is proper, $L(\lambda)\ne 0$. We prove that $L(\lambda)$ is irreducible. Let $W\subset L(\lambda)$ be a $\mathfrak g$-submodule, and define its inverse image under $\pi$ by \begin{align*} \widetilde W:=\pi^{-1}(W)\subset M(\lambda). \end{align*} Then $\widetilde W$ is a $\mathfrak g$-submodule containing $J(\lambda)$. If $W\ne L(\lambda)$, then $\widetilde W\ne M(\lambda)$, so $\widetilde W$ is proper. By maximality of $J(\lambda)$, we get $\widetilde W=J(\lambda)$, and therefore $W=0$. Thus $L(\lambda)$ has no proper nonzero submodules, so it is irreducible. [/step] [step:Show the quotient is the unique irreducible highest weight quotient of $M(\lambda)$] Because $v_\lambda\notin J(\lambda)$, the vector \begin{align*} \bar v_\lambda:=\pi(v_\lambda)\in L(\lambda) \end{align*} is nonzero. For every $h\in\mathfrak h$ and $x\in\mathfrak n^+$, \begin{align*} h\bar v_\lambda=\lambda(h)\bar v_\lambda \end{align*} and \begin{align*} x\bar v_\lambda=0, \end{align*} because these identities hold for $v_\lambda$ in $M(\lambda)$ and $\pi$ is a $\mathfrak g$-[module homomorphism](/page/Module%20Homomorphism). Since $M(\lambda)=U(\mathfrak g)v_\lambda$, applying $\pi$ gives \begin{align*} L(\lambda)=U(\mathfrak g)\bar v_\lambda. \end{align*} Thus $L(\lambda)$ is a highest weight module of highest weight $\lambda$. Finally, let $Q$ be any irreducible quotient of $M(\lambda)$, and let $\rho:M(\lambda)\to Q$ be a surjective $\mathfrak g$-module homomorphism. Its kernel $\ker\rho$ is a proper submodule because $Q\ne 0$, so $\ker\rho\subset J(\lambda)$. Since $Q$ is irreducible, $\ker\rho$ is maximal among proper submodules: if $\ker\rho\subset K\subset M(\lambda)$, then $K/\ker\rho$ is a submodule of $Q$, hence either $0$ or $Q$. Therefore $K=\ker\rho$ or $K=M(\lambda)$. By uniqueness of the maximal proper submodule, $\ker\rho=J(\lambda)$. Hence $Q\cong M(\lambda)/J(\lambda)=L(\lambda)$ as $\mathfrak g$-modules. This proves that $L(\lambda)$ is the unique irreducible quotient of $M(\lambda)$ and completes the proof. [/step]