Let me fix an infinite-dimensional (complex) Banach space $E$. There is a cute result of Bracic and Kuzma which says that every maximal abelian subalgebra of $\mathscr{B}(E)$, the algebra of bounded operators of $E$ is infinite-dimensional. Nice!

Let us take then an abelian closed subalgebra (feel free to take a maximal one, if you wish) $\mathscr{A}\subset \mathscr{B}(E)$ and a proper ideal $\mathscr{I}$ of $\mathscr{A}$. Now, consider the the left ideal $\mathscr{L}$ of $\mathscr{B}(E)$ generated by $\mathscr{I}$. May it happen that $\mathscr{L}$ is improper, that is, $\mathscr{L}=\mathscr{B}(E)$? Of course, for general rings such situation is fully legitimate.