[proofplan]
We prove the contrapositive in both directions. If $L$ decomposes as a nontrivial direct sum of ideals, then the Cartan subalgebra and root spaces decompose accordingly, and the Killing form makes roots from different summands orthogonal. Conversely, if the root system splits into two nonempty orthogonal parts, we build two Lie subalgebras from the corresponding root spaces and Cartan directions; the root-space bracket relations show these subalgebras commute and form a nontrivial direct-sum decomposition of $L$ into ideals.
[/proofplan]
[step:Record the root-space structure used in the proof]
Let $\kappa: L \times L \to k$ denote the Killing form of $L$. Since $L$ is semisimple, Cartan's criterion gives that $\kappa$ is nondegenerate on $L$. We use the standard structural theorem for finite-dimensional semisimple Lie algebras over algebraically closed fields of characteristic zero, namely the [Root-space decomposition and root-string relations for semisimple Lie algebras](/theorems/???). This theorem also states that the restriction $\kappa|_{H \times H}$ is nondegenerate. Hence, for each $\lambda \in H^*$, define $\tau(\lambda) \in H$ to be the unique element satisfying
\begin{align*}
\kappa(\tau(\lambda),h)=\lambda(h)
\end{align*}
for every $h \in H$.
We use the following consequences of the same structural theorem:
\begin{align*}
L = H \oplus \bigoplus_{\alpha \in \Phi} L_\alpha,
\end{align*}
the subspace $H$ is spanned by $\{\tau(\alpha):\alpha \in \Phi\}$, and for roots $\alpha,\beta \in \Phi$,
\begin{align*}
[L_\alpha,L_\beta] \subset
\begin{cases}
L_{\alpha+\beta}, & \alpha+\beta \in \Phi,\\
H, & \alpha+\beta = 0,\\
0, & \alpha+\beta \notin \Phi \cup \{0\}.
\end{cases}
\end{align*}
More precisely, for each $\alpha \in \Phi$,
\begin{align*}
[L_\alpha,L_{-\alpha}]=k\tau(\alpha),
\qquad
(\alpha,\alpha)\neq 0.
\end{align*}
The symmetric [bilinear form](/page/Bilinear%20Form) on $\operatorname{span}_{\mathbb{R}}\Phi$ is given by
\begin{align*}
(\alpha,\beta) = \kappa(\tau(\alpha),\tau(\beta)).
\end{align*}
Equivalently, for every $\alpha,\beta \in \Phi$,
\begin{align*}
\beta(\tau(\alpha))=(\alpha,\beta).
\end{align*}
[guided]
Let $\kappa: L \times L \to k$ be the Killing form. The semisimplicity of $L$ is used first through Cartan's criterion: $\kappa$ is nondegenerate on $L$. A nondegenerate bilinear form on a [vector space](/page/Vector%20Space) need not restrict nondegenerately to every subspace, so we need an additional structural input for the Cartan subalgebra. The [Root-space decomposition and root-string relations for semisimple Lie algebras](/theorems/???) states, under the present hypotheses that $L$ is finite-dimensional semisimple over an algebraically closed field of characteristic zero and $H$ is a Cartan subalgebra, that $\kappa|_{H \times H}$ is nondegenerate. Therefore $\kappa|_{H \times H}$ identifies $H$ with $H^*$. Thus, for every $\lambda \in H^*$, there is a unique element $\tau(\lambda) \in H$ such that
\begin{align*}
\kappa(\tau(\lambda),h)=\lambda(h)
\end{align*}
for all $h \in H$.
We also use the standard structural facts for roots of semisimple Lie algebras from the same theorem. The algebra decomposes as
\begin{align*}
L = H \oplus \bigoplus_{\alpha \in \Phi} L_\alpha,
\end{align*}
where
\begin{align*}
L_\alpha = \{x \in L : [h,x]=\alpha(h)x \text{ for every } h \in H\}.
\end{align*}
The Cartan subalgebra is spanned by the root directions $\tau(\alpha)$:
\begin{align*}
H=\operatorname{span}_k\{\tau(\alpha):\alpha \in \Phi\}.
\end{align*}
Finally, the bracket of two root spaces is controlled by addition of roots:
\begin{align*}
[L_\alpha,L_\beta] \subset
\begin{cases}
L_{\alpha+\beta}, & \alpha+\beta \in \Phi,\\
H, & \alpha+\beta = 0,\\
0, & \alpha+\beta \notin \Phi \cup \{0\}.
\end{cases}
\end{align*}
The same structural theorem gives the sharper relation
\begin{align*}
[L_\alpha,L_{-\alpha}]=k\tau(\alpha)
\end{align*}
for each $\alpha \in \Phi$, and it gives non-isotropy of roots:
\begin{align*}
(\alpha,\alpha)\neq 0.
\end{align*}
These are the algebraic rules that allow us to translate a decomposition of $\Phi$ into a decomposition of $L$.
The inner product on the real root space $\operatorname{span}_{\mathbb{R}}\Phi$ is induced by the Killing form:
\begin{align*}
(\alpha,\beta)=\kappa(\tau(\alpha),\tau(\beta)).
\end{align*}
By the defining property of $\tau(\beta)$, this also gives
\begin{align*}
\beta(\tau(\alpha))=(\alpha,\beta).
\end{align*}
This identity is what turns orthogonality of roots into vanishing of Cartan actions.
[/guided]
[/step]
[step:Show that a nontrivial ideal decomposition makes the root system reducible]
Assume $L$ is not simple. Since $L$ is semisimple, there exist nonzero ideals $I,J \trianglelefteq L$ such that
\begin{align*}
L = I \oplus J.
\end{align*}
For each $x \in L$, let
\begin{align*}
\operatorname{ad}x: L &\to L \\
z &\mapsto [x,z]
\end{align*}
denote the adjoint endomorphism of $L$. The Killing form satisfies $\kappa(I,J)=0$, because for $x \in I$ and $y \in J$ the operators $\operatorname{ad}x$ and $\operatorname{ad}y$ have images in $I$ and $J$ respectively, and the ideals commute:
\begin{align*}
[I,J] \subset I \cap J = 0.
\end{align*}
Let
\begin{align*}
H_I = H \cap I,
\qquad
H_J = H \cap J.
\end{align*}
By the componentwise Cartan decomposition for a direct sum of semisimple ideals, applied to the decomposition $L=I\oplus J$ and the Cartan subalgebra $H\subset L$, a Cartan subalgebra decomposes as
\begin{align*}
H = H_I \oplus H_J,
\end{align*}
with $H_I$ and $H_J$ Cartan subalgebras of $I$ and $J$, respectively. The same result says that each root space of $L$ relative to $H$ lies entirely in exactly one summand.
Define
\begin{align*}
\Phi_I &= \{\alpha \in \Phi : L_\alpha \subset I\},\\
\Phi_J &= \{\alpha \in \Phi : L_\alpha \subset J\}.
\end{align*}
The root-space decomposition of $L=I\oplus J$ gives
\begin{align*}
\Phi = \Phi_I \sqcup \Phi_J,
\end{align*}
and both sets are nonempty because $I$ and $J$ are nonzero semisimple Lie algebras.
If $\alpha \in \Phi_I$ and $\beta \in \Phi_J$, then $\tau(\alpha) \in H_I$ and $\tau(\beta) \in H_J$. Since $\kappa(I,J)=0$,
\begin{align*}
(\alpha,\beta)=\kappa(\tau(\alpha),\tau(\beta))=0.
\end{align*}
Thus $\Phi$ is the union of two nonempty mutually orthogonal subsets, so $\Phi$ is reducible.
[guided]
Suppose $L$ is not simple. Because $L$ is semisimple, every ideal has a complementary semisimple ideal, so there are nonzero ideals $I,J \trianglelefteq L$ with
\begin{align*}
L=I\oplus J.
\end{align*}
The two ideals commute. Indeed,
\begin{align*}
[I,J]\subset I
\end{align*}
because $I$ is an ideal, and also
\begin{align*}
[I,J]\subset J
\end{align*}
because $J$ is an ideal. Hence
\begin{align*}
[I,J]\subset I\cap J=0.
\end{align*}
Now we compare the Killing form on the two summands. For each $x \in L$, define the adjoint endomorphism
\begin{align*}
\operatorname{ad}x: L &\to L \\
z &\mapsto [x,z].
\end{align*}
If $x \in I$ and $y \in J$, then $\operatorname{ad}x$ maps $I$ into $I$ and kills $J$ after composition with $\operatorname{ad}y$, while $\operatorname{ad}y$ maps $J$ into $J$ and kills $I$ after composition with $\operatorname{ad}x$. Thus $\operatorname{ad}x \operatorname{ad}y$ has trace zero, so
\begin{align*}
\kappa(x,y)=0.
\end{align*}
Therefore
\begin{align*}
\kappa(I,J)=0.
\end{align*}
Let
\begin{align*}
H_I=H\cap I,
\qquad
H_J=H\cap J.
\end{align*}
We use the componentwise Cartan decomposition for direct sums of semisimple ideals. Its hypotheses are satisfied because $I$ and $J$ are semisimple ideals and $L=I\oplus J$. It gives
\begin{align*}
H=H_I\oplus H_J.
\end{align*}
The same result says that the roots of $L$ relative to $H$ are exactly the roots coming from $I$ and the roots coming from $J$, extended by zero on the other Cartan summand, and that each root space lies entirely in the corresponding ideal. Accordingly define
\begin{align*}
\Phi_I &= \{\alpha \in \Phi : L_\alpha \subset I\},\\
\Phi_J &= \{\alpha \in \Phi : L_\alpha \subset J\}.
\end{align*}
Then
\begin{align*}
\Phi=\Phi_I\sqcup\Phi_J.
\end{align*}
Both $\Phi_I$ and $\Phi_J$ are nonempty, since $I$ and $J$ are nonzero semisimple Lie algebras and therefore have nonempty root systems relative to their Cartan subalgebras.
It remains to check orthogonality. Choose $\alpha \in \Phi_I$ and $\beta \in \Phi_J$. The Killing-dual elements satisfy
\begin{align*}
\tau(\alpha)\in H_I,
\qquad
\tau(\beta)\in H_J.
\end{align*}
Since $H_I \subset I$ and $H_J \subset J$, and since $\kappa(I,J)=0$, we get
\begin{align*}
(\alpha,\beta)=\kappa(\tau(\alpha),\tau(\beta))=0.
\end{align*}
Thus every root from $\Phi_I$ is orthogonal to every root from $\Phi_J$. Hence $\Phi$ is reducible.
[/guided]
[/step]
[step:Build Lie subalgebras from an orthogonal decomposition of the root system]
Assume conversely that $\Phi$ is reducible. Then there are nonempty subsets $\Phi_1,\Phi_2 \subset \Phi$ such that
\begin{align*}
\Phi=\Phi_1\sqcup\Phi_2
\end{align*}
and
\begin{align*}
(\alpha,\beta)=0
\end{align*}
for every $\alpha \in \Phi_1$ and every $\beta \in \Phi_2$.
For $i \in \{1,2\}$, define the Cartan part
\begin{align*}
H_i=\operatorname{span}_k\{\tau(\alpha):\alpha \in \Phi_i\}\subset H
\end{align*}
and define the subspace
\begin{align*}
L_i = H_i \oplus \bigoplus_{\alpha \in \Phi_i} L_\alpha \subset L.
\end{align*}
Because $\Phi_i$ is closed under negation and under addition whenever the sum is a root, the root-space bracket relations imply
\begin{align*}
[L_i,L_i]\subset L_i.
\end{align*}
Thus $L_1$ and $L_2$ are Lie subalgebras of $L$.
[guided]
Now suppose the root system is reducible. This means that we can write it as a disjoint union
\begin{align*}
\Phi=\Phi_1\sqcup\Phi_2
\end{align*}
where both $\Phi_1$ and $\Phi_2$ are nonempty and mutually orthogonal:
\begin{align*}
(\alpha,\beta)=0
\end{align*}
for all $\alpha \in \Phi_1$ and $\beta \in \Phi_2$.
First observe that each $\Phi_i$ is closed under negation. If $\alpha \in \Phi_1$, then $-\alpha$ cannot lie in $\Phi_2$, because
\begin{align*}
(\alpha,-\alpha)=-(\alpha,\alpha)\neq 0.
\end{align*}
Hence $-\alpha \in \Phi_1$. The same argument applies to $\Phi_2$.
Next, each $\Phi_i$ is closed under addition when the sum is a root. Suppose $\alpha,\beta \in \Phi_1$ and $\alpha+\beta \in \Phi$. If $\alpha+\beta$ lay in $\Phi_2$, then it would be orthogonal to both $\alpha$ and $\beta$. Therefore
\begin{align*}
(\alpha+\beta,\alpha+\beta)
=
(\alpha+\beta,\alpha)+(\alpha+\beta,\beta)
=
0+0
=
0,
\end{align*}
which is impossible for a root. Hence $\alpha+\beta \in \Phi_1$. The same argument applies to $\Phi_2$.
For $i \in \{1,2\}$, define
\begin{align*}
H_i=\operatorname{span}_k\{\tau(\alpha):\alpha \in \Phi_i\}\subset H
\end{align*}
and
\begin{align*}
L_i=H_i\oplus \bigoplus_{\alpha \in \Phi_i}L_\alpha.
\end{align*}
We verify that $L_i$ is a Lie subalgebra. Since $H$ is abelian,
\begin{align*}
[H_i,H_i]=0.
\end{align*}
If $h \in H_i$ and $x_\alpha \in L_\alpha$ with $\alpha \in \Phi_i$, then
\begin{align*}
[h,x_\alpha]=\alpha(h)x_\alpha \in L_\alpha \subset L_i.
\end{align*}
Finally, if $x_\alpha \in L_\alpha$ and $x_\beta \in L_\beta$ with $\alpha,\beta \in \Phi_i$, then the root-space bracket relations give
\begin{align*}
[x_\alpha,x_\beta]\subset
\begin{cases}
L_{\alpha+\beta}, & \alpha+\beta \in \Phi,\\
H_i, & \alpha+\beta=0,\\
0, & \alpha+\beta \notin \Phi\cup\{0\}.
\end{cases}
\end{align*}
The first case lies in $L_i$ because $\Phi_i$ is closed under root sums, and the second case lies in $H_i$ because $[L_\alpha,L_{-\alpha}]$ is contained in the Cartan direction generated by $\tau(\alpha)$. Hence
\begin{align*}
[L_i,L_i]\subset L_i.
\end{align*}
So $L_i$ is a Lie subalgebra.
[/guided]
[/step]
[step:Show that the two subalgebras commute]
We claim that
\begin{align*}
[L_1,L_2]=0.
\end{align*}
First, since $H$ is abelian,
\begin{align*}
[H_1,H_2]=0.
\end{align*}
If $h=\tau(\alpha)\in H_1$ with $\alpha \in \Phi_1$ and $x_\beta \in L_\beta$ with $\beta \in \Phi_2$, then
\begin{align*}
[h,x_\beta]=\beta(h)x_\beta=\beta(\tau(\alpha))x_\beta=(\alpha,\beta)x_\beta=0.
\end{align*}
By linearity, $[H_1,L_\beta]=0$ for every $\beta \in \Phi_2$, and similarly $[H_2,L_\alpha]=0$ for every $\alpha \in \Phi_1$.
Finally, let $\alpha \in \Phi_1$ and $\beta \in \Phi_2$. Since $\alpha+\beta$ cannot be a root and $\alpha+\beta \neq 0$, the root-space bracket relation gives
\begin{align*}
[L_\alpha,L_\beta]=0.
\end{align*}
Therefore $[L_1,L_2]=0$.
[guided]
We now prove that the two subalgebras commute. The Cartan parts commute because $H$ is abelian:
\begin{align*}
[H_1,H_2]=0.
\end{align*}
Next consider the action of one Cartan part on the other root spaces. Let $h=\tau(\alpha)$ for some $\alpha \in \Phi_1$, and let $x_\beta \in L_\beta$ for some $\beta \in \Phi_2$. By the definition of a root space,
\begin{align*}
[h,x_\beta]=\beta(h)x_\beta.
\end{align*}
Using the definition of $\tau(\alpha)$ and the induced root form,
\begin{align*}
\beta(h)=\beta(\tau(\alpha))=(\alpha,\beta).
\end{align*}
The two root families are orthogonal, so
\begin{align*}
(\alpha,\beta)=0.
\end{align*}
Thus
\begin{align*}
[h,x_\beta]=0.
\end{align*}
Linearity gives $[H_1,L_\beta]=0$ for every $\beta \in \Phi_2$. The same argument with the roles of $\Phi_1$ and $\Phi_2$ reversed gives $[H_2,L_\alpha]=0$ for every $\alpha \in \Phi_1$.
It remains to bracket root spaces from different components. Let $\alpha \in \Phi_1$ and $\beta \in \Phi_2$. We first show that $\alpha+\beta$ is not a root. If $\alpha+\beta \in \Phi_1$, then it must be orthogonal to $\beta \in \Phi_2$, but
\begin{align*}
(\alpha+\beta,\beta)=(\alpha,\beta)+(\beta,\beta)=0+(\beta,\beta)\neq 0.
\end{align*}
If $\alpha+\beta \in \Phi_2$, then it must be orthogonal to $\alpha \in \Phi_1$, but
\begin{align*}
(\alpha+\beta,\alpha)=(\alpha,\alpha)+(\beta,\alpha)=(\alpha,\alpha)+0\neq 0.
\end{align*}
Both possibilities are impossible. Also $\alpha+\beta \neq 0$, because $\beta=-\alpha$ would force $\beta$ to lie in the same component as $\alpha$. Therefore
\begin{align*}
\alpha+\beta \notin \Phi\cup\{0\}.
\end{align*}
The root-space bracket relation gives
\begin{align*}
[L_\alpha,L_\beta]=0.
\end{align*}
Combining the Cartan-root and root-root computations proves
\begin{align*}
[L_1,L_2]=0.
\end{align*}
[/guided]
[/step]
[step:Conclude that reducibility of the root system gives a nontrivial ideal decomposition]
Since $H$ is spanned by $\{\tau(\alpha):\alpha \in \Phi\}$ and $\Phi=\Phi_1\sqcup\Phi_2$, we have
\begin{align*}
H=H_1+H_2.
\end{align*}
The orthogonality of $\Phi_1$ and $\Phi_2$ implies $H_1\cap H_2=0$, because an element of $H_1\cap H_2$ is orthogonal to every $\tau(\alpha)$ with $\alpha \in \Phi$, hence orthogonal to all of $H$, and the Killing form is nondegenerate on $H$. Thus
\begin{align*}
H=H_1\oplus H_2.
\end{align*}
Using the root-space decomposition,
\begin{align*}
L=L_1\oplus L_2.
\end{align*}
Both $L_1$ and $L_2$ are nonzero because $\Phi_1$ and $\Phi_2$ are nonempty. Since $[L_1,L_2]=0$ and each $L_i$ is a Lie subalgebra, each $L_i$ is an ideal of $L$. Therefore $L$ is not simple.
Combining this with the previous direction, $L$ is simple if and only if $\Phi$ is irreducible.
[guided]
We finish by proving that the vector-space sum constructed from the two orthogonal root components is a direct-sum decomposition of $L$ by ideals. The structural theorem recorded at the start gives
\begin{align*}
H=\operatorname{span}_k\{\tau(\alpha):\alpha\in\Phi\}.
\end{align*}
Since $\Phi=\Phi_1\sqcup\Phi_2$ and
\begin{align*}
H_i=\operatorname{span}_k\{\tau(\alpha):\alpha\in\Phi_i\}
\end{align*}
for $i\in\{1,2\}$, we get
\begin{align*}
H=H_1+H_2.
\end{align*}
We next show that this sum is direct. Let $h\in H_1\cap H_2$. Because $h\in H_1$, it is a $k$-linear combination of elements $\tau(\alpha)$ with $\alpha\in\Phi_1$; because every root in $\Phi_2$ is orthogonal to every root in $\Phi_1$, we have
\begin{align*}
\kappa(h,\tau(\beta))=0
\end{align*}
for every $\beta\in\Phi_2$. Similarly, because $h\in H_2$, we have
\begin{align*}
\kappa(h,\tau(\alpha))=0
\end{align*}
for every $\alpha\in\Phi_1$. Hence $h$ is orthogonal to every element of $\{\tau(\gamma):\gamma\in\Phi\}$. Since these elements span $H$, $h$ is orthogonal to all of $H$. The restriction $\kappa|_{H\times H}$ is nondegenerate by the structural theorem recorded at the start, so $h=0$. Therefore
\begin{align*}
H=H_1\oplus H_2.
\end{align*}
Combining this with the root-space decomposition
\begin{align*}
L=H\oplus\bigoplus_{\alpha\in\Phi}L_\alpha
\end{align*}
and the disjoint union $\Phi=\Phi_1\sqcup\Phi_2$ gives
\begin{align*}
L=L_1\oplus L_2.
\end{align*}
Each $L_i$ is nonzero because $\Phi_i$ is nonempty and root spaces are nonzero by definition of a root. To show that $L_i$ is an ideal of $L$, use the decomposition $L=L_1\oplus L_2$. Since $L_i$ is already a Lie subalgebra, $[L_i,L_i]\subset L_i$. Since the preceding step proved $[L_1,L_2]=0$, the bracket of $L_i$ with the other summand is also contained in $L_i$. Therefore
\begin{align*}
[L,L_i]\subset L_i,
\end{align*}
so each $L_i$ is an ideal of $L$. The decomposition is nontrivial, so $L$ is not simple.
The first direction proved that a nontrivial ideal decomposition makes $\Phi$ reducible, and this direction proved that reducibility of $\Phi$ makes $L$ non-simple. Thus $L$ is simple if and only if $\Phi$ is irreducible.
[/guided]
[/step]
