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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.

edited Dec 4 2011 at 13:56

asked Dec 4 2011 at 13:09

Sellapan Nathan
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Will the algebra $K(H)$ of compact operators on a Hilbet space satisfy you? It has not non-trivial (closed) two-sided ideals. – Sergei Akbarov Dec 4 2011 at 13:33

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@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?? – Johannes Hahn Dec 4 2011 at 13:34

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Perhaps the Calkin algebrea $B(H)/K(H)$ is a good candidate. – Sellapan Nathan Dec 4 2011 at 13:53

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Right, Sellapan. There are many Banach algebras that have a largest (necessarily closed) ideal which you can mod out to get an example. – Bill Johnson Dec 4 2011 at 16:48

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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). – Yemon Choi Dec 4 2011 at 21:08

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Alain Valette · Answer 1 · 2011-12-04 16:03:08Z UTC

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

answered Dec 4 2011 at 16:03

Alain Valette
6,439●14●26

Algebraically simple Banach algebras

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