Are maximal ideals in C*-algebras closed?

Surprisingly, the following question is open:

Question: Are maximal ideals in C*-algebras closed?

Let $A$ be a C*-algebra. By an ideal in $A$ we mean a two-sided ideal $I \subseteq A$ that is not necessarily closed. We say that $I$ is a maximal ideal if the only ideals $J \subseteq A$ satisfying $I \subseteq J$ are $I$ and $A$ . An ideal $I$ is proper if $I \neq A$ .

If $A$ is unital, then every maximal ideal is closed. Indeed, in this case, using that a proper ideal does not contain any invertible element of $A$ , and using that the set of invertible elements is open, we see that if $I \subseteq A$ is a proper ideal, then so is its closure $\overline{I}$ . Thus, if $I$ is maximal, then $I = \overline{I}$ , and $I$ is closed.

Further, it is known that every maximal left ideal in a C*-algebra is closed; see for example Proposition 3.5 in [1]. (This even holds in Banach algebras admitting a bounded approximate unit.) In commutative C*-algebras, every maximal ideal is also a maximal left ideal and therefore closed.

Assume $I$ is a non-closed, maximal ideal in a C*-algebra $A$ (if it exists). Then $I$ is dense and therefore contains the Pederson ideal of $A$ . Further, the quotient $A/I$ is a simple $\mathbb{C}$ -algebra. It is easy to see that $A/I$ is radical, that is, $\mathrm{rad}(A/I)=A/I$ , because $A/I$ has no nonzero maximal (modular) left ideals. One can show that $I$ is hereditary (if $0 \leq x \leq y$ in $A$ , and $y$ belongs to $I$ , then so does $x$ ), strongly invariant (if $xx^* \in I$ , then $x^*x \in I$ ), and invariant under powers: if $a \in I_+$ and $t > 0$ , then $a^t \in I$ . For the arguments see this post in mathoverflow.

[1]
M.C. García, H.G. Dales, Á.R. Palacios, Maximal left ideals in Banach algebras, Bull. London Math. Soc. (2019) 1–15. https://doi.org/10.1112/blms.12290.

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