# Is it true that for simple

Is it true that for simple ${C}^{\ast }$-algebras, meaning that they don't have non-trivial two-sided ideals, it holds that they are necessarily non-commutative? And why?
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Nola Rivera
This is false. $\mathbb{C}$ is commutative and simple.
But this is the only simple commutative ${C}^{\ast }$-algebra. This is because if $A$ is not isomorphic to $\mathbb{C}$, then it has a non-zero, non-invertible element $x$ by the Gelfand-Mazur theorem. If $A$ is also commutative, then $Ax$ is a proper ideal.

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Theresa Archer
For a different point of view, a commutative $C\ast$-algebra $A$ is isomorphic, via the Gelfand transform, to ${C}_{0}\left(X\right)$ for a locally compact Hausdorff space $X$. If $X$ is a singleton, then $A=\mathbb{C}$. Otherwise, for nontrivial closed $Y\subset X$, then set

is a closed two-sided ideal.

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