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At a primitive cube root of unity, $[1]=1$, $[2]=-1$, $[3]=0$. In the degree-three quantum plane module, put $z_0=x^3$, $z_1=x^{2}y$, $z_2=x{y}^2$, $z_3=y^3$. The complete actions areHence $W=⟨z_0,z_3⟩$ is a two-dimensional trivial submodule; every line in $W$ is a simple module. The quotient by $W$ is a two-dimensional simple module, with distinct $K$ eigenvalues $q,q^{-1}$ and nonzero maps connecting its two weight spaces.

Let $S$ be a simple module that is a submodule of this cubic component. If $S\cap W\ne 0$, simplicity makes $S\subseteq W$, so $S$ is one of its lines. If $S\cap W=0$, the quotient map injects $S$ into the simple quotient. Its nonzero image is the entire quotient, so $\dim S=2$. Since $K$ is diagonalizable, such an $S$ must contain the unique $q$ weight space $\mathbb{C}{z}_1$. But $E{z}_1=z_0\ne 0$, which then belongs to $S\cap W$, a contradiction. Therefore the cubic-root quantum-plane socle gives the complete answer:All simple submodules lie in the two-dimensional $W$, so their sum cannot equal the four-dimensional cubic component. It is not a direct sum of simple modules. The obstruction is the nonzero extension to the simple two-dimensional quotient, not a failure of that quotient to be simple.