[proofplan] We argue by contradiction from the definition of simplicity. If $\mathfrak g$ decomposed as a direct sum of two nonzero ideals, then either summand would be a nonzero ideal of $\mathfrak g$. The direct-sum condition forces each summand to be proper, contradicting the assumption that a simple Lie algebra has no nonzero proper ideals. [/proofplan] [step:Assume a nontrivial direct-sum decomposition by ideals] Suppose, for contradiction, that there exist ideals $I,J \trianglelefteq \mathfrak g$ with $I \neq 0$, $J \neq 0$, and \begin{align*} \mathfrak g = I \oplus J. \end{align*} Here $\oplus$ means that $\mathfrak g = I + J$ as a $k$-[vector space](/page/Vector%20Space) and \begin{align*} I \cap J = \{0\}. \end{align*} [/step] [step:Show that one summand is a nonzero proper ideal] By assumption, $I$ is a nonzero ideal of $\mathfrak g$. We claim that $I \neq \mathfrak g$. If $I = \mathfrak g$, then since $J \subseteq \mathfrak g = I$, we would have \begin{align*} J \subseteq I \cap J = \{0\}, \end{align*} so $J = 0$, contradicting the assumption that $J \neq 0$. Hence $I$ is a nonzero proper ideal of $\mathfrak g$. [/step] [step:Contradict simplicity and conclude indecomposability] Since $\mathfrak g$ is simple, its only ideals are $0$ and $\mathfrak g$. The preceding step produced an ideal $I \trianglelefteq \mathfrak g$ satisfying \begin{align*} I \neq 0 \qquad\text{and}\qquad I \neq \mathfrak g, \end{align*} which contradicts simplicity. Therefore no such nonzero ideals $I$ and $J$ exist, and $\mathfrak g$ is indecomposable. [/step]