I'm trying to come up with an example of a Banach algebra $A$ that is not commutative, unital and such that the only closed ideals are $\{0\}$ and $A$.
I already struggled to even come up with a non abelian Banach algebra. The only two examples I could think of were matrix groups and the space of bounded linear oeprators. (are there any more?)
From there I went on to think about matrix groups since it seemed to promise a simpler example. I was trying to use the determinant as an isomorphism: if you start with the general linear group and note that the special linear group is a subgroup (they're all rings, really) then the determinant is a surjective homomorphism from the general linear group into $\mathbb C \setminus \{0\}$. To make it into an isomorphism one can quotient the general linear group by the special linear group (the special linear group is the kernel). But the problem with this idea is that once the domain is a quotient (set of equivalence classes) the map doesn't seem to be well-defined anymore: if $A$ has determinant $a$ and $B$ has determinant $1$ then $\mathrm{det}(A + B)$ should equal $\mathrm{det}(A)$ since the two matrices differ by an element of the special linear group.
On the other hand, since $\mathrm{det}$ is a homomorphism, $\mathrm{det}(A + B)$ should equal $\mathrm{det}(A)+\mathrm{det}(B) = \mathrm{det}(A)+1$.
What am I doing wrong here?
Then I started to wonder about why they should be closed ideals. I know that a maximal ideal in a Banach algebra is closed but the converse does probably not hold. So what's the significance of closedness here?
Edit Apparently I was on the right track. If you look here at example 3.1.1. then matrix algebras can be show to have no nonzero proper two sided ideals. Would that be enough for an example or can a matrix algebra still have a one sided closed proper ideal?