The derived affine algebra omits only the degree derivation

Jacobi makes $[\mathfrak{g},\mathfrak{g}]$ an ideal. It is nonzero because $\mathfrak{g}$ is nonabelian, so simplicity gives $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$. Write any $x\in \mathfrak{g}$ as a finite sum ${\sum }_j[y_j,z_j]$. For any $m\in \mathbb{Z}$, ${\sum }_j[(y_j{)}_m,(z_j{)}_0]=x_m$: the central coefficient is $m{\delta }_{m,0}=0$. Thus every loop mode, including degree zero, lies in the derived algebra, using only brackets within $\overline{L\mathfrak{g}}$.

Choose $h,h^{\mathrm{\prime }}\in \mathfrak{h}$ with $B(h,h^{\mathrm{\prime }})=1$, possible by F2 and rescaling the nondegenerate form. The Cartan is nonzero since its roots span the dual and a nonzero simple algebra cannot have an empty root decomposition. Then $[h_1,h_{-1}^{\mathrm{\prime }}]=c$, because $[h,h^{\mathrm{\prime }}]=0$. Hence $c$ also lies in the derived algebra. This uses just two finite-dimensional vector choices, not AC.

Every defining bracket has zero $d$ coordinate. Bilinearity gives $[\widehat{\mathfrak{g}},\widehat{\mathfrak{g}}]\subseteq L\mathfrak{g}\oplus \mathbb{C}c$, while steps 1.1–1.2 give the reverse inclusion. Since $d$ is a separate nonzero basis vector, this subspace is proper. The same two steps also show the central extension itself is perfect.