There are plenty of semi-simple Banach algebras - this broad class includes C*-algebras and algebras of bounded operators on a given Banach space. On the other hand, it seems unlikely to me that there exists (infinite-dimensional) algebraically simple Banach algebra, that is, a Banach algebra $A$ such that if $J\subseteq A$ is a two-sided ideal, then either $J=\{0\}$ or $J=A$. Is my conjecture true?

EDIT: Of course, the Calkin algebra is the answer. You can delete my question.

algebraicallysimple. In other words: Are there non-closed two-sided ideals other than {0} and the whole algebra?? $\endgroup$ – Johannes Hahn Dec 4 '11 at 13:34commutativeBanach algebra (necessarily non-unital and radical) which is algebraically simple? The question has been open since the 1970s (I think). $\endgroup$ – Yemon Choi Dec 4 '11 at 21:08