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This is a problem in exercise one of Murphy's book

Find an example of a nonabelian unital Banach algebra $A$, where the only closed ideals are $\{0\}$ and $A$.

But does such an algebra exist at all?

My argument is the following:

Let $a$ be an arbitrary nonzero element in $A$, if $a$ is not invertible, then $a$ is contained in a maximal ideal, which is closed. However, there is no such ideal thus every nonzero element must be invertible. Then Gelfand-Mazur says $A$ is the complex numbers and thus must be abelian.

What is the problem with my argument?


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up vote 4 down vote accepted

The problem is that not being invertible doesn't imply being contained in a maximal two-sided ideal if $A$ is noncommutative. For example, there are plenty of elements of $\mathcal{M}_n(\mathbb{C})$ that are not invertible but since this algebra is simple (which incidentally implies that it's already an answer to your question when $n \ge 2$) all nonzero elements generate the unit two-sided ideal.

What is true is that not being left-invertible is equivalent to being contained in a maximal left ideal and not being right-invertible is equivalent to being contained in a maximal right ideal.

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Thanks, Qiaochu! – Hui Yu Jun 29 '12 at 1:57

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