# Dimension of quotient of ideals

Let $A$ be a Banach algebra, $I$ be a closed two-sided ideal in $A$, and $J$ be a closed two-sided ideal in $I$ such that there is no ideal between $I$ and $J$. Can we see $dim(\frac{I}{J})<\infty$?

Y. DOMAR in "On the ideal structure of certain Banach algebras", proves a lemma like this question with the following difference:

$A$ is commutative, $J$ is a two-sided ideal in $A$ and he shows $dim(\frac{I}{J})=1$

• Do you know about the ideal structure of B(H)? This provides an immediate noncommutative counterexample. One place to learn more is by looking at some of the articles cited in www.math.uni.wroc.pl/~drygier/ivmrt2014/slides/laustsen.pdf – Yemon Choi Jan 10 '17 at 15:58
• What about above question when $A$ is commutative? – Albert harold Jan 10 '17 at 16:29
• $\mathbb{C}_p$ is a commutative Banach algebra over $\mathbb{Q}_p$ and one can set $J=0$ and $I=\mathbb{C}_p$ for a counterexample in the commutative case... – Kevin Buzzard Jan 10 '17 at 18:13
• Oh rotten luck :-) What is this pathological non-non-archimedean complex numbers field anyway? Is there an infinite complete field extension of it? That would presumably be an example in the commutative case. Can you just take some random transcendental extension and then complete it somehow? – Kevin Buzzard Jan 10 '17 at 20:04
• @KevinBuzzard In the land of complex Banach algebras, the only one which is also a field is C itself (Gelfand-Mazur theorem). The second idea seems to lead towards a notorious open problem: does there exist a unital commutative Banach algebra over C with no non-trivial proper closed ideals? – Yemon Choi Jan 10 '17 at 22:31

If we allow $A$ to be noncommutative then there are well-known natural counterexamples. For instance one can look at $A=B(E)$ for various Banach spaces $E$. I imagine that there should be counterexamples of the form $A=L^1(G)$ but I do not know this for sure.
There are "silly" commutative counterexamples obtained by defining $I$ to be an infinite-dimensional Banach space with zero product, $J=\{0\}$, and setting $A$ to be the unitization of $I$.