Let $g$ a simple Lie algebra with root space decomposition $g = \mathfrak{h} \oplus \bigoplus_{\alpha \in \phi} g_{\alpha}$, where $\mathfrak{h}$ is a Cartan subalgebra and $\phi$ is the root system. Let $\mathfrak{b} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \phi^+} g_{\alpha}$ a Borel subalgebra of $g$. I want to prove that if $I$ is an abelian ideal of $\mathfrak{b}$, then $I = \bigoplus_{\alpha \in \psi} g_{\alpha}$ with $\psi \subseteq \phi^+$.
I tried this approach: let $x \in I - \mathfrak{h}$, which exists because if $I = \mathfrak{h}$ and $e_{\alpha} \in g_{\alpha}$, $[x, e_{\alpha}] = \alpha(x)e_{\alpha} \in I$, which is a contradiction. So now it is enough to prove that $\mathfrak{h} \cap I = \left\{ 0 \right\}$. By contradiction, let $y \in \mathfrak{h} \cap I$. Then $[x, y] = 0$, which is a contradiction since $C_g(\mathfrak{h}) = \mathfrak{h}$. Is this correct?
Now, knowing that $I$ is an abelian ideal of $\mathfrak{b}$ iff $\psi$ is a subset of $\phi^{+}$ such that if $\alpha, \beta \in \psi$, then $\alpha + \beta \notin \phi^{+}$, I want to describe all the abelian ideals of $\mathfrak{b}$ in the case of $g = \mathfrak{sl}(4)$. I thought of describing them in terms of $\psi$. So I obtained 8 possible ideals, each one with a root system $\psi$ made of two roots of type $\epsilon_i - \epsilon_j$ such that their sum does not belong to $\phi^{+}$. However I don’t think it’s totally correct to assume that the root system of each ideal has always order two. What am I missing? Is there another description of those ideals?