Assume that $\mathfrak{g}$ is a finite-dimensional semisimple Lie algebra over an algebraically closed field $F$ of characteristic $0, \mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ and $\Delta$ is its root system. Linear map $\nu: \mathfrak{h} \rightarrow \mathfrak{h}^{*}$ is an isomorphism defined by $(\nu(h))\left(h^{\prime}\right)=$ $K\left(h, h^{\prime}\right)$ for $h, h^{\prime} \in \mathfrak{h}$.
Proposition : If $\Delta$ is decomposable, i.e. $\Delta=\Delta^{\prime} \cup \Delta^{\prime \prime}$, with $\alpha^{\prime}+\alpha^{\prime \prime} \notin$ $\Delta \cup\{0\}$ for all $\alpha^{\prime} \in \Delta, \alpha^{\prime \prime} \in \Delta$ then we have a corresponding decomposition of $\mathfrak{g}$ into direct sum of ideals $\mathfrak{g}=\mathfrak{g}^{\prime} \oplus \mathfrak{g}^{\prime \prime}$ where $$ \mathfrak{g}^{\prime}=\mathfrak{h}^{\prime} \oplus\left(\bigoplus_{\alpha^{\prime} \in \Delta^{\prime}} \mathfrak{g}_{\alpha^{\prime}}\right) \\ $$ $ \text { for } \mathfrak{h}^{\prime}=\operatorname{span}\left\{\nu^{-1}\left(\Delta^{\prime}\right)\right\} \text { and } \\ $ $$ \mathfrak{g}^{\prime \prime}=\mathfrak{h}^{\prime \prime} \oplus\left(\bigoplus_{\alpha^{\prime \prime} \in \Delta^{\prime \prime}} \mathfrak{g}_{\alpha^{\prime \prime}}\right) \\ $$ $ \text { for } \mathfrak{h}^{\prime \prime}=\operatorname{span}\left\{\nu^{-1}\left(\Delta^{\prime \prime}\right)\right\} . $
Proof: Because $\alpha^{\prime}+\alpha^{\prime \prime} \notin \Delta \cup\{0\}$, we get, $\left[\mathfrak{g}_{\alpha^{\prime}}, \mathfrak{g}_{\alpha^{\prime \prime}}\right] \subseteq \mathfrak{g}_{\alpha^{\prime}+\alpha^{\prime \prime}}=0$. Since $\sum_{\alpha \in \Delta^{\prime}} F\left[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}\right]=F \nu^{-1}\left(\Delta^{\prime}\right)$ we conclude, say by Jacobi identity, $\left[\mathfrak{h}^{\prime}, \mathfrak{g}_{\alpha^{\prime \prime}}\right]=0$ and $\left[\mathfrak{h}^{\prime \prime}, \mathfrak{g}_{\alpha^{\prime}}\right]=0$, so $\left[\mathfrak{g}^{\prime}, \mathfrak{g}^{\prime \prime}\right]=0$. It remains to show that $\mathfrak{g}^{\prime}$ and $\mathfrak{g}^{\prime \prime}$ are subalgebras. First note that $\beta^{\prime} \in \Delta^{\prime}$ implies $-\beta^{\prime} \in \Delta^{\prime}$. Indeed, we have $-\beta^{\prime} \in \Delta$, and supposing $-\beta^{\prime} \in \Delta^{\prime \prime}$ leads to $\beta^{\prime}+\left(-\beta^{\prime}\right) \in \Delta \cup\{0\}$, a contradiction. Now if $\alpha^{\prime}, \beta^{\prime} \in \Delta^{\prime}$ then $\gamma=\alpha^{\prime}+\beta^{\prime} \notin \Delta^{\prime \prime}$, because that would mean $\gamma+\left(-\beta^{\prime}\right)=\alpha^{\prime} \in \Delta$ for $\gamma \in \Delta^{\prime \prime}$ and $-\beta^{\prime} \in \Delta^{\prime}$, another contradiction. So either $\alpha^{\prime}, \beta^{\prime} \in \Delta^{\prime} \cup 0$ and $\left[\mathfrak{g}_{\alpha^{\prime}}, \mathfrak{g}_{\beta^{\prime}}\right]=\mathfrak{g}_{\alpha^{\prime}+\beta^{\prime}} \in \mathfrak{g}^{\prime}$ or $\alpha^{\prime}, \beta^{\prime} \notin \Delta \cup 0$ and $\left[\mathfrak{g}_{\alpha^{\prime}}, \mathfrak{g}_{\beta^{\prime}}\right]=0 .$ It is now clear that $\mathfrak{g}^{\prime}$ is a subalgebra. Similarly, $\mathfrak{g}^{\prime \prime}$ is also a subalgebra.
My question is :
(1) In first line of proof, I think '$F \nu^{-1}\left(\Delta^{\prime}\right)$' should be repalce by '$\operatorname{span}\left\{\nu^{-1}\left(\Delta^{\prime}\right)\right\}$', because $\sum_{\alpha \in \Delta^{\prime}} F\left[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}\right]$ is a linear combination. Is that right?
(2) Why $\mathfrak{g}^{\prime} + \mathfrak{g}^{\prime \prime}$ is direct sum? Does $\mathfrak{h}=\mathfrak{h}^{\prime} \oplus \mathfrak{h}^{\prime \prime}$?
Thanks for your help.