# Example of a Banach algebra $A$ whose only closed ideals are $\{ 0 \}$ and $A$.

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?

• Quick comment: The determinant is only a group homomorhism, NOT an algebra-homomorphim. That's why you can't factor out $\rm{SL}$. – PhoemueX May 29 '14 at 13:06
• Matrix groups aren't algebras... – Branimir Ćaćić May 30 '14 at 16:57
• @Student The ring of $2\times2$ matrices over $\mathbb{C}$ has no ideal other than $\{0\}$ and the ring itself; since the operator norm makes it into a Banach algebra… – egreg May 31 '14 at 12:45
• @Student: Your comment that $\mathcal{B}(\mathcal{H})$, the algebra of bounded linear operators on a Hilbert space $\mathcal{H}$, has no proper non-trivial closed ideals is not generally true. If $\mathcal{H}$ is infinite-dimensional, then it contains $\mathcal{K}(\mathcal{H})$, the algebra of compact operators, as a proper non-trivial closed ideal. If $\mathcal{H}$ is finite-dimensional, then $\mathcal{B}(\mathcal{H}) = \mathcal{K}(\mathcal{H})$, which I believe that you already know. – Berrick Caleb Fillmore Jun 9 '14 at 23:29
• @Student: Dear Student, You consistently write "matrix groups" in these comments, but matrix groups are not Banach algebras (or any kind of algebra). To get rings of matrices you have to look at the collection of all operators (on some appropriate vector space), not just the invertible ones. It might be a good idea to clear up this confusion. Regards, – Matt E Jun 10 '14 at 20:52

For each $n \in \mathbb{N}$, the matrix algebra ${\mathbb{M}_{n}}(\mathbb{C})$ has indeed no non-trivial proper ideals. A proof of this is as follows.

Proof: For notational convenience, let $[n] = \{ 1,\ldots,n \}$. For each $(i,j) \in [n] \times [n]$, let $I_{i j}$ denote the $(n \times n)$-matrix with $(i,j)$-th entry $1$ and $0$’s elsewhere. Then observe that $$\forall (i,j),(\alpha,\beta) \in [n] \times [n]: \quad I_{i j} I_{\alpha \beta} = \delta_{j \alpha} I_{i \beta},$$ where $\delta$ is Kronecker’s delta. For the sake of contradiction, suppose that there exists a non-trivial proper ideal $J$ of ${\mathbb{M}_{n}}(\mathbb{C})$. Pick a non-zero element $A \in J$, and write $A$ as a linear combination of the $I_{i j}$’s: $$A = \sum_{i,j = 1}^{n} a_{i j} I_{i j}, \quad \text{where  a_{i j} \in \mathbb{C} .}$$ Then $a_{r s} \neq 0$ for some $(r,s) \in [n] \times [n]$. As $I_{r s} A I_{s r} = a_{r s} I_{r r} \in J$ (check!), we have $I_{r r} \in J$. Hence, $I_{i j} = I_{i r} I_{r r} I_{r j} \in J$ for all $(i,j) \in [n] \times [n]$, which yields $J = {\mathbb{M}_{n}}(\mathbb{C})$. Contradiction. $\quad \spadesuit$

There are many other examples of Banach algebras with no non-trivial closed proper ideals. Here are a few well-known $C^{*}$-algebraic examples (we call them simple $C^{*}$-algebras).

• The algebra of compact linear operators on a Hilbert space $\mathcal{H}$, denoted by $\mathcal{K}(\mathcal{H})$. (This is a standard example and is mentioned by Branimir Ćaćić in his comment above. It subsumes the example of ${\mathbb{M}_{n}}(\mathbb{C})$, since ${\mathbb{M}_{n}}(\mathbb{C}) \cong \mathcal{B}(\mathbb{C}^{n}) = \mathcal{K}(\mathbb{C}^{n})$.)

• For each irrational $\theta \in \mathbb{R}$, the irrational rotation algebra $A_{\theta}$. This is the universal $C^{*}$-algebra generated by unitaries $u$ and $v$ subject to the relation $u v = e^{2 \pi i \theta} v u$. (To learn more about $A_{\theta}$, please consult the paper $C^{*}$-Algebras Associated with Irrational Rotations by Marc Rieffel.)

• For each $n \in \mathbb{N}_{\geq 2}$, the Cuntz algebra $\mathcal{O}_{n}$. This is the universal $C^{*}$-algebra generated by $n$ isometries $S_{1},\ldots,S_{n}$ subject to the relation $\displaystyle \sum_{i = 1}^{n} S_{i} S_{i}^{*} = 1$. (To learn more about $\mathcal{O}_{n}$, please consult the paper Simple $C^{*}$-Algebras Generated by Isometries by Joachim Cuntz.)

• The reduced group $C^{*}$-algebra ${C^{*}_{r}}(G * H)$, where $G$ and $H$ are discrete groups not both of order $2$. (This example is due to William Paschke and Norberto Salinas, and it is the subject of their paper $C^{*}$-Algebras Associated with Free Products of Groups.)

Once you know some simple $C^{*}$-algebras, you can use the following result to create others.

Theorem (Takesaki): If $A$ and $B$ are simple $C^{*}$-algebras, then so is $A \otimes_{\min} B$.

Unfortunately, I cannot think of non-$C^{*}$-algebraic examples, but I hope that this answer suffices. Good luck with your search!

• Thank you for this wonderful answer. The proof is beautiful! And you give me even more information than I ask for. – Student Jun 19 '14 at 17:59