## Subalgebras of $B(E)$

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.

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## 1 Answer

(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 $\mathscr{B}(E)$.

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 Looks good to me, Phil. This suggests that Tomek's question should really be about proper ideals in maximal abelian subalgebras... – Yemon Choi Sep 30 at 2:40 Oh, yes. That was easy but I am interested in masas indeed. Let met accept this answer within a week if nobody else attacks the 'masa' case. – Tomek Kania Oct 1 at 17:53