[proofplan] The proof uses [Lie's theorem](/theorems/3754): a finite-dimensional representation of a solvable complex Lie algebra has a common eigenvector. Applying this to the given $\mathfrak g$-module $V$ produces a nonzero vector whose one-dimensional span is stable under the action of every element of $\mathfrak g$. Since $V$ is simple, this nonzero submodule must be all of $V$, so $V$ is one-dimensional. [/proofplan] [step:Apply Lie's theorem to obtain a common eigenvector] Let \begin{align*} \rho: \mathfrak g &\to \mathfrak{gl}_{\mathbb C}(V) \end{align*} be the representation defining the $\mathfrak g$-module structure on $V$, so that $X \cdot v = \rho(X)v$ for $X \in \mathfrak g$ and $v \in V$. Since $\mathfrak g$ is finite-dimensional and solvable over $\mathbb C$, and since $V$ is a nonzero finite-dimensional complex [vector space](/page/Vector%20Space), [Lie's theorem](/theorems/3802) applies to the representation $\rho$ (citing a result not yet in the wiki: [Lie's Theorem](/theorems/3803)). Hence there exist a nonzero vector $v_0 \in V$ and a linear functional \begin{align*} \lambda: \mathfrak g &\to \mathbb C \end{align*} such that \begin{align*} \rho(X)v_0 = \lambda(X)v_0 \end{align*} for every $X \in \mathfrak g$. [guided] The key input is Lie's theorem, which says that a finite-dimensional representation of a solvable Lie algebra over an algebraically closed field has a common eigenvector. Here the field is $\mathbb C$, which is algebraically closed, the Lie algebra $\mathfrak g$ is solvable by hypothesis, and the module $V$ is nonzero and finite-dimensional by hypothesis. Therefore Lie's theorem applies to the representation \begin{align*} \rho: \mathfrak g &\to \mathfrak{gl}_{\mathbb C}(V). \end{align*} The conclusion is stronger than saying that each operator $\rho(X)$ has some eigenvector. It gives a single nonzero vector $v_0 \in V$ that is an eigenvector for every operator $\rho(X)$ at once. Thus there is a linear functional \begin{align*} \lambda: \mathfrak g &\to \mathbb C \end{align*} such that, for every $X \in \mathfrak g$, \begin{align*} \rho(X)v_0 = \lambda(X)v_0. \end{align*} This common eigenvector is exactly what is needed to construct a one-dimensional submodule. [/guided] [/step] [step:Show that the common eigenspace line is a nonzero submodule] Define the one-dimensional complex subspace \begin{align*} L := \mathbb C v_0 \subseteq V. \end{align*} Since $v_0 \neq 0$, we have $L \neq \{0\}$. We verify that $L$ is a $\mathfrak g$-submodule of $V$. Let $X \in \mathfrak g$ and let $w \in L$. Then there exists $a \in \mathbb C$ such that $w = a v_0$. Using linearity of $\rho(X): V \to V$ and the eigenvector relation, \begin{align*} X \cdot w &= \rho(X)(a v_0) \\ &= a \rho(X)v_0 \\ &= a \lambda(X)v_0 \\ &\in \mathbb C v_0 \\ &= L. \end{align*} Thus $X \cdot L \subseteq L$ for every $X \in \mathfrak g$, so $L$ is a nonzero $\mathfrak g$-submodule of $V$. [/step] [step:Use simplicity to force the submodule to be all of $V$] Because $V$ is simple as a $\mathfrak g$-module, its only $\mathfrak g$-submodules are $\{0\}$ and $V$. The submodule $L$ is nonzero, so $L = V$. Therefore \begin{align*} \dim_{\mathbb C} V = \dim_{\mathbb C} L = 1. \end{align*} This proves that every nonzero finite-dimensional simple $\mathfrak g$-module is one-dimensional. [/step]