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In the finite-dimensional setting of the preceding part, let $S$ be a simple module. Each action map $s\mapsto as$ is an $A$-module endomorphism because $A$ is commutative. The complex Schur's lemma makes every such action a scalar. Therefore every linear subspace of $S$ is an submodule; simplicity forcesEquivalently, a simple module over a commutative ring is a quotient by a maximal ideal, and a finite-dimensional field extension of $\mathbb{C}$ is just $\mathbb{C}$.

The finite-dimensional context matters. If the clause is read as allowing an arbitrary commutative complex algebra, the unrestricted assertion is false: $A=\mathbb{C}(t)$ is simple as a module over itself and is infinite-dimensional over $\mathbb{C}$. This is simple module of a finite-dimensional commutative complex algebra. Thus the one-dimensional conclusion uses the finite-dimensional algebra context, or an alternative hypothesis such as finite dimensionality of the simple module.