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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.

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  • $\begingroup$ Will the algebra $K(H)$ of compact operators on a Hilbet space satisfy you? It has not non-trivial (closed) two-sided ideals. $\endgroup$
    – Sergei Akbarov
    Dec 4, 2011 at 13:33
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    $\begingroup$ @Sergei: I think, that's exactly the point: is this algebraically simple. In other words: Are there non-closed two-sided ideals other than {0} and the whole algebra?? $\endgroup$
    – Johannes Hahn
    Dec 4, 2011 at 13:34
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    $\begingroup$ Perhaps the Calkin algebrea $B(H)/K(H)$ is a good candidate. $\endgroup$
    – Sellapan Nathan
    Dec 4, 2011 at 13:53
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    $\begingroup$ Right, Sellapan. There are many Banach algebras that have a largest (necessarily closed) ideal which you can mod out to get an example. $\endgroup$
    – Bill Johnson
    Dec 4, 2011 at 16:48
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    $\begingroup$ It may be worth remarking that a notorious open question in Banach algebras asks: does there exist an infinite-dimensional commutative Banach 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, 2011 at 21:08

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Observe that, since the group of invertible elements is open, any simple unital Banach algebra is algebraically simple.

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