[proofplan] We use the PBW basis of a Verma module to write every vector uniquely as a finite linear combination of $f^k v_\lambda$. The commutation relations give an explicit formula for $e f^k v_\lambda$, and this formula identifies exactly when a lower-weight basis vector is singular. A minimal-degree argument then shows that every nonzero submodule contains a singular vector. This proves simplicity unless $\lambda$ is a nonnegative integer, and in the exceptional case identifies the unique submodule generated by $f^{n+1}v_n$ and the two composition factors. [/proofplan] [step:Write the Verma module in its PBW basis] By the Poincare-Birkhoff-Witt theorem [citetheorem:8827], applied to the ordered basis $f,h,e$ of $\mathfrak{sl}_2(\mathbb C)$, the Verma module $M(\lambda)$ has the $\mathbb C$-basis \begin{align*} \{f^k v_\lambda:k\in\mathbb Z_{\ge 0}\}. \end{align*} For each $k\in\mathbb Z_{\ge 0}$, the commutation relation $[h,f]=-2f$ gives \begin{align*} h f^k v_\lambda=(\lambda-2k)f^k v_\lambda. \end{align*} Thus $f^k v_\lambda$ has weight $\lambda-2k$, and every element of $M(\lambda)$ has a unique expression \begin{align*} w=\sum_{k=0}^{m} c_k f^k v_\lambda \end{align*} with $m\in\mathbb Z_{\ge 0}$ and coefficients $c_0,\dots,c_m\in\mathbb C$. [/step] [step:Compute the singular-vector condition] We prove that for every integer $k\ge 1$, \begin{align*} e f^k v_\lambda=k(\lambda-k+1)f^{k-1}v_\lambda. \end{align*} For $k=1$, the relation $[e,f]=h$ gives \begin{align*} e f v_\lambda=f e v_\lambda+h v_\lambda=\lambda v_\lambda. \end{align*} Assume the formula holds for some $k\ge 1$. Since $h f^k v_\lambda=(\lambda-2k)f^k v_\lambda$, we get \begin{align*} e f^{k+1}v_\lambda=f e f^k v_\lambda+h f^k v_\lambda. \end{align*} Substituting the induction hypothesis gives \begin{align*} e f^{k+1}v_\lambda=k(\lambda-k+1)f^k v_\lambda+(\lambda-2k)f^k v_\lambda. \end{align*} The coefficient is \begin{align*} k(\lambda-k+1)+(\lambda-2k)=(k+1)(\lambda-k). \end{align*} Therefore \begin{align*} e f^{k+1}v_\lambda=(k+1)(\lambda-k)f^k v_\lambda. \end{align*} This completes the induction. Consequently, for $k\ge 1$, the vector $f^k v_\lambda$ is killed by $e$ if and only if \begin{align*} k(\lambda-k+1)=0. \end{align*} Since $k\ne 0$, this is equivalent to $\lambda=k-1$. Thus a singular basis vector below $v_\lambda$ occurs exactly when $\lambda\in\mathbb Z_{\ge 0}$, in which case the unique such index is $k=\lambda+1$. [/step] [step:Show every nonzero submodule contains a singular basis vector] Let $N\subseteq M(\lambda)$ be a nonzero $\mathfrak{sl}_2(\mathbb C)$-submodule. Choose a nonzero element \begin{align*} w=\sum_{k=a}^{b} c_k f^k v_\lambda\in N \end{align*} such that $a,b\in\mathbb Z_{\ge 0}$, $a\le b$, $c_a\ne 0$, $c_b\ne 0$, and $a$ is minimal among the lowest occurring indices of nonzero elements of $N$. Because $N$ is stable under $h$, the vector \begin{align*} \prod_{\substack{a\le j\le b, j\ne a}}(h-(\lambda-2j))w \end{align*} belongs to $N$. Since the weights $\lambda-2j$ are distinct for distinct $j$, this vector is a nonzero scalar multiple of $f^a v_\lambda$. Hence $f^a v_\lambda\in N$. If $a\ge 1$ and $e f^a v_\lambda\ne 0$, then the formula from the previous step gives a nonzero scalar multiple of $f^{a-1}v_\lambda$ in $N$, contradicting the minimality of $a$. Therefore either $a=0$, or $e f^a v_\lambda=0$. In particular, if $N$ is proper, then $a\ne 0$, because $v_\lambda\in N$ would imply $N=M(\lambda)$. Hence every nonzero proper submodule contains a singular basis vector $f^a v_\lambda$ with $a\ge 1$. [guided] The point of this step is to replace the informal phrase “choose a vector of maximal weight” by a finite and completely controlled argument. A submodule of a Verma module may be infinite-dimensional, so we instead choose a nonzero vector whose lowest $f$-degree is as small as possible. Let $N\subseteq M(\lambda)$ be a nonzero $\mathfrak{sl}_2(\mathbb C)$-submodule. Every nonzero vector of $M(\lambda)$ has a unique finite expansion in the PBW basis, so choose \begin{align*} w=\sum_{k=a}^{b} c_k f^k v_\lambda\in N \end{align*} with $a,b\in\mathbb Z_{\ge 0}$, $a\le b$, $c_a\ne 0$, $c_b\ne 0$, and with $a$ minimal among all lowest occurring indices of nonzero vectors in $N$. We now extract the lowest-degree term by using the diagonal action of $h$. For each $j$, the vector $f^j v_\lambda$ has $h$-weight $\lambda-2j$. These weights are distinct because $\lambda-2i=\lambda-2j$ implies $i=j$. Therefore the polynomial in $h$ \begin{align*} \prod_{\substack{a\le j\le b, j\ne a}}(h-(\lambda-2j)) \end{align*} annihilates every term $f^jv_\lambda$ with $j\ne a$ and multiplies $f^a v_\lambda$ by the nonzero scalar \begin{align*} \prod_{\substack{a\le j\le b, j\ne a}}((\lambda-2a)-(\lambda-2j)). \end{align*} Since $N$ is a submodule, it is stable under the action of $h$ and hence under every polynomial in $h$. Thus the vector \begin{align*} \prod_{\substack{a\le j\le b, j\ne a}}(h-(\lambda-2j))w \end{align*} belongs to $N$ and is a nonzero scalar multiple of $f^a v_\lambda$. Hence $f^a v_\lambda\in N$. Now suppose $a\ge 1$ and $e f^a v_\lambda\ne 0$. Since $N$ is stable under $e$, the vector $e f^a v_\lambda$ belongs to $N$. By the action formula, \begin{align*} e f^a v_\lambda=a(\lambda-a+1)f^{a-1}v_\lambda. \end{align*} Under the present assumption this is a nonzero scalar multiple of $f^{a-1}v_\lambda$, contradicting the minimality of $a$. Therefore either $a=0$ or $e f^a v_\lambda=0$. Finally, if $N$ is proper, then $a$ cannot be $0$. Indeed, $f^0v_\lambda=v_\lambda$, and the highest weight vector $v_\lambda$ generates the Verma module by construction, so $v_\lambda\in N$ would force $N=M(\lambda)$. Thus every nonzero proper submodule contains a singular vector $f^a v_\lambda$ with $a\ge 1$. [/guided] [/step] [step:Deduce simplicity outside the nonnegative integers] Assume $\lambda\notin\mathbb Z_{\ge 0}$. If $N\subseteq M(\lambda)$ were a nonzero proper submodule, the previous step would give a singular vector $f^a v_\lambda\in N$ with $a\ge 1$. The singular-vector calculation shows that this requires $\lambda=a-1\in\mathbb Z_{\ge 0}$, a contradiction. Hence no nonzero proper submodule exists, and $M(\lambda)$ is simple. [/step] [step:Identify the unique proper submodule when $\lambda=n$] Now let $n\in\mathbb Z_{\ge 0}$. The action formula gives \begin{align*} e f^{n+1}v_n=(n+1)(n-(n+1)+1)f^n v_n=0. \end{align*} Moreover \begin{align*} h f^{n+1}v_n=(n-2n-2)f^{n+1}v_n=(-n-2)f^{n+1}v_n. \end{align*} Thus $f^{n+1}v_n$ is a highest weight vector of weight $-n-2$. Let \begin{align*} N_n:=U(\mathfrak{sl}_2(\mathbb C))\, f^{n+1}v_n\subseteq M(n) \end{align*} be the submodule it generates. By the [universal property of Verma modules](/theorems/9375) [citetheorem:9375], there is a unique $\mathfrak{sl}_2(\mathbb C)$-[module homomorphism](/page/Module%20Homomorphism) \begin{align*} \Phi:M(-n-2)\to N_n \end{align*} sending $v_{-n-2}$ to $f^{n+1}v_n$. The map $\Phi$ is surjective by the definition of $N_n$. Since $-n-2\notin\mathbb Z_{\ge 0}$, the simplicity already proved implies that $M(-n-2)$ is simple. The map $\Phi$ is nonzero, so its kernel is not all of $M(-n-2)$; simplicity forces $\ker\Phi=0$. Hence $\Phi$ is an isomorphism, and \begin{align*} N_n\cong M(-n-2). \end{align*} If $N\subseteq M(n)$ is any nonzero proper submodule, the minimal-degree argument gives a singular vector $f^a v_n\in N$ with $a\ge 1$. The singular-vector calculation shows that $a=n+1$. Therefore $f^{n+1}v_n\in N$, so $N_n\subseteq N$. Conversely, $N$ cannot contain any $f^a v_n$ with $a\le n$, because applying $e$ repeatedly would eventually place $v_n$ in $N$, forcing $N=M(n)$. Hence $N\subseteq N_n$. Thus $N=N_n$, and $N_n$ is the unique nonzero proper submodule. [/step] [step:Compute the composition factors] The quotient $M(n)/N_n$ has basis represented by \begin{align*} v_n,fv_n,\dots,f^n v_n. \end{align*} It is generated by the image of $v_n$, which has highest weight $n$. Any nonzero submodule of this quotient contains, by the same minimal-degree argument, a nonzero singular vector represented by some $f^a v_n$ with $0\le a\le n$. For $1\le a\le n$, the action formula gives \begin{align*} e f^a v_n=a(n-a+1)f^{a-1}v_n\ne 0. \end{align*} Thus the only singular vector in the quotient is the highest weight line spanned by the image of $v_n$, and any nonzero submodule contains it. Since that image generates the quotient, the quotient is simple. Therefore \begin{align*} M(n)/N_n\cong L(n). \end{align*} Since $N_n\cong M(-n-2)$ and $M(-n-2)$ is simple, we also have \begin{align*} N_n\cong L(-n-2). \end{align*} The short filtration \begin{align*} 0\subset N_n\subset M(n) \end{align*} therefore has simple factors $L(-n-2)$ and $L(n)$, each occurring exactly once. This proves the claimed list of composition factors and completes the proof. [/step]