[proofplan] The first assertion follows by applying irreducibility to the kernel and image of a nonzero homomorphism: the kernel is a submodule of the source, and the image is a submodule of the target. The second assertion applies the first assertion to an endomorphism after subtracting an eigenvalue. Algebraic closedness and finite-dimensionality guarantee an eigenvalue, and the corresponding eigenspace gives a nonzero kernel for $S - \lambda \operatorname{id}_V$. [/proofplan] [step:Show that the kernel and image of a module homomorphism are submodules] Let $T: V \to W$ be a $\mathfrak g$-module homomorphism. Define \begin{align*} \ker T &= \{v \in V : T(v) = 0\}, \\ \operatorname{im} T &= \{T(v) : v \in V\}. \end{align*} Both are $k$-linear subspaces because $T$ is $k$-linear. We verify stability under the $\mathfrak g$-action. If $x \in \mathfrak g$ and $v \in \ker T$, then the module homomorphism property gives \begin{align*} T(x \cdot v) = x \cdot T(v) = x \cdot 0 = 0, \end{align*} so $x \cdot v \in \ker T$. Thus $\ker T$ is a $\mathfrak g$-submodule of $V$. If $w \in \operatorname{im} T$, choose $v \in V$ with $w = T(v)$. For every $x \in \mathfrak g$, \begin{align*} x \cdot w = x \cdot T(v) = T(x \cdot v), \end{align*} again by the module homomorphism property. Hence $x \cdot w \in \operatorname{im} T$, so $\operatorname{im} T$ is a $\mathfrak g$-submodule of $W$. [guided] Let $T: V \to W$ be a $\mathfrak g$-module homomorphism. This means that $T$ is $k$-linear and intertwines the two $\mathfrak g$-actions: \begin{align*} T(x \cdot v) = x \cdot T(v) \end{align*} for every $x \in \mathfrak g$ and every $v \in V$. Define \begin{align*} \ker T &= \{v \in V : T(v) = 0\}, \\ \operatorname{im} T &= \{T(v) : v \in V\}. \end{align*} The $k$-linearity of $T$ implies that $\ker T$ is a $k$-linear subspace of $V$ and $\operatorname{im} T$ is a $k$-linear subspace of $W$. To prove that these are submodules, we must also check stability under the action of every element of $\mathfrak g$. First take $v \in \ker T$ and $x \in \mathfrak g$. Since $T(v) = 0$, the homomorphism property gives \begin{align*} T(x \cdot v) = x \cdot T(v) = x \cdot 0 = 0. \end{align*} Thus $x \cdot v \in \ker T$, so $\ker T$ is stable under the $\mathfrak g$-action. Therefore $\ker T$ is a $\mathfrak g$-submodule of $V$. Now take $w \in \operatorname{im} T$ and $x \in \mathfrak g$. By definition of the image, there exists $v \in V$ such that $w = T(v)$. Then \begin{align*} x \cdot w = x \cdot T(v) = T(x \cdot v). \end{align*} The final expression lies in $\operatorname{im} T$, because $x \cdot v \in V$. Hence $\operatorname{im} T$ is stable under the $\mathfrak g$-action, so $\operatorname{im} T$ is a $\mathfrak g$-submodule of $W$. [/guided] [/step] [step:Use irreducibility to prove that a nonzero homomorphism is an isomorphism] Assume now that $T: V \to W$ is nonzero. Since $\ker T$ is a $\mathfrak g$-submodule of the irreducible module $V$, irreducibility gives either $\ker T = 0$ or $\ker T = V$. The second possibility would imply $T = 0$, contrary to hypothesis. Hence $\ker T = 0$, so $T$ is injective. Similarly, $\operatorname{im} T$ is a $\mathfrak g$-submodule of the irreducible module $W$. Since $T \ne 0$, there exists $v \in V$ such that $T(v) \ne 0$, so $\operatorname{im} T \ne 0$. Irreducibility of $W$ therefore gives $\operatorname{im} T = W$, so $T$ is surjective. Thus $T$ is a bijective $\mathfrak g$-module homomorphism, hence an isomorphism of $\mathfrak g$-modules. [guided] Assume that $T: V \to W$ is a nonzero $\mathfrak g$-module homomorphism. From the previous step, $\ker T$ is a $\mathfrak g$-submodule of $V$. Since $V$ is irreducible, its only $\mathfrak g$-submodules are $0$ and $V$. Therefore \begin{align*} \ker T = 0 \quad \text{or} \quad \ker T = V. \end{align*} If $\ker T = V$, then every $v \in V$ satisfies $T(v) = 0$, so $T$ is the zero map. This contradicts the hypothesis that $T \ne 0$. Hence $\ker T = 0$, and therefore $T$ is injective. Again from the previous step, $\operatorname{im} T$ is a $\mathfrak g$-submodule of $W$. Because $T$ is not the zero map, there exists $v \in V$ such that $T(v) \ne 0$. Thus $\operatorname{im} T$ is not the zero submodule. Since $W$ is irreducible, the only nonzero $\mathfrak g$-submodule of $W$ is $W$ itself, so \begin{align*} \operatorname{im} T = W. \end{align*} Thus $T$ is surjective. We have shown that $T$ is both injective and surjective. Since $T$ is already a $\mathfrak g$-module homomorphism, it is an isomorphism of $\mathfrak g$-modules. [/guided] [/step] [step:Produce an eigenvalue for a finite-dimensional endomorphism over an algebraically closed field] Assume that $k$ is algebraically closed and that $V$ is finite-dimensional over $k$. Let $S: V \to V$ be a $\mathfrak g$-module endomorphism. Since $V$ is irreducible, $V \ne 0$; set $n := \dim_k V$, so $n \ge 1$. Choose a $k$-basis of $V$, and let $A \in k^{n \times n}$ be the matrix of the $k$-[linear map](/page/Linear%20Map) $S$ in this basis. Define the characteristic polynomial \begin{align*} p_S: k \to k, \qquad \lambda \mapsto \det(A - \lambda I_n). \end{align*} This polynomial has degree $n \ge 1$. Since $k$ is algebraically closed, there exists $\lambda \in k$ such that \begin{align*} p_S(\lambda) = 0. \end{align*} Thus $\det(A - \lambda I_n) = 0$, so the linear map $S - \lambda \operatorname{id}_V$ has nonzero kernel. [guided] Now assume that $k$ is algebraically closed and that $V$ is finite-dimensional over $k$. Let $S: V \to V$ be a $\mathfrak g$-module endomorphism. We first need an eigenvalue of the underlying $k$-linear map. This is the only point where finite-dimensionality and algebraic closedness are used. Since $V$ is irreducible, it is nonzero. Define \begin{align*} n := \dim_k V. \end{align*} Then $n \ge 1$. Choose a $k$-basis of $V$, and let $A \in k^{n \times n}$ be the matrix representing the $k$-linear map $S$ with respect to this basis. Define the characteristic polynomial \begin{align*} p_S: k \to k, \qquad \lambda \mapsto \det(A - \lambda I_n). \end{align*} This polynomial has degree $n$, hence positive degree. Since $k$ is algebraically closed, every nonconstant polynomial over $k$ has a root in $k$. Therefore there exists $\lambda \in k$ such that \begin{align*} p_S(\lambda) = 0. \end{align*} By the definition of $p_S$, this means \begin{align*} \det(A - \lambda I_n) = 0. \end{align*} A square matrix has determinant zero exactly when the corresponding linear map is not injective. Therefore the $k$-linear map \begin{align*} S - \lambda \operatorname{id}_V: V \to V \end{align*} has nonzero kernel. Equivalently, $\lambda$ is an eigenvalue of $S$. [/guided] [/step] [step:Apply irreducibility to the eigenspace and conclude scalarity] The map \begin{align*} S - \lambda \operatorname{id}_V: V \to V \end{align*} is a $\mathfrak g$-module endomorphism: for every $x \in \mathfrak g$ and every $v \in V$, \begin{align*} (S - \lambda \operatorname{id}_V)(x \cdot v) &= S(x \cdot v) - \lambda x \cdot v \\ &= x \cdot S(v) - x \cdot (\lambda v) \\ &= x \cdot (S(v) - \lambda v) \\ &= x \cdot (S - \lambda \operatorname{id}_V)(v). \end{align*} By the previous step, $\ker(S - \lambda \operatorname{id}_V) \ne 0$. Since this kernel is a $\mathfrak g$-submodule of the irreducible module $V$, irreducibility gives \begin{align*} \ker(S - \lambda \operatorname{id}_V) = V. \end{align*} Therefore $(S - \lambda \operatorname{id}_V)(v) = 0$ for every $v \in V$, so \begin{align*} S = \lambda \operatorname{id}_V. \end{align*} This proves that every $\mathfrak g$-module endomorphism of $V$ is multiplication by a scalar in $k$. [guided] We have found $\lambda \in k$ such that the $k$-linear map \begin{align*} S - \lambda \operatorname{id}_V: V \to V \end{align*} has nonzero kernel. To use irreducibility, we must verify that this map is still a $\mathfrak g$-module homomorphism. Take $x \in \mathfrak g$ and $v \in V$. Since $S$ is a $\mathfrak g$-module endomorphism, it satisfies \begin{align*} S(x \cdot v) = x \cdot S(v). \end{align*} Using also the $k$-bilinearity of the module action, we compute \begin{align*} (S - \lambda \operatorname{id}_V)(x \cdot v) &= S(x \cdot v) - \lambda x \cdot v \\ &= x \cdot S(v) - x \cdot (\lambda v) \\ &= x \cdot (S(v) - \lambda v) \\ &= x \cdot (S - \lambda \operatorname{id}_V)(v). \end{align*} Thus $S - \lambda \operatorname{id}_V$ is a $\mathfrak g$-module endomorphism of $V$. By the kernel argument from the first step, $\ker(S - \lambda \operatorname{id}_V)$ is a $\mathfrak g$-submodule of $V$. The previous step showed that this kernel is nonzero. Since $V$ is irreducible, any nonzero submodule of $V$ must be all of $V$, hence \begin{align*} \ker(S - \lambda \operatorname{id}_V) = V. \end{align*} This equality means that every $v \in V$ satisfies \begin{align*} (S - \lambda \operatorname{id}_V)(v) = 0. \end{align*} Equivalently, \begin{align*} S(v) = \lambda v \end{align*} for every $v \in V$. Therefore \begin{align*} S = \lambda \operatorname{id}_V. \end{align*} So every $\mathfrak g$-module endomorphism of $V$ is scalar multiplication by an element of $k$. [/guided] [/step]