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.



Subalgebras of $B(E)$
(Thinking and writing about this in a hurry, so take it with a grain of salt!).


Let $E$ be the space $\ell\_p(\mathbb{Z})$, $1\leq p\leq \infty$, and $T$ the left (or right) shift operator on $E$. Let $\mathscr{A}$ be the (abelian) closed subalgebra of $\mathscr{B}(E)$ generated by $T$ and $\mathscr{I}$ be the closed subalgebra of $\mathscr{B}(E)$ generated by the set of powers of the form $ T^n$, $n\geq 2$. Then $\mathscr{I}$ is a proper ideal of $\mathscr{A}$ and, since it contains the invertible operator $T^2$, the left ideal $\mathscr{L}$ coincides with $\mathscr{B}(E)$.


