# Can we embed unital Banach algebras into semi-simple ones?

A Banach algebra is (Jacobson) semi-simple if the intersection of all maximal left ideals is the zero ideal.

Take a unital abelian Banach algebra $B$. Can we embed it unitally into an abelian semi-simple Banach algebra?

My attempt was to take the algebra $A=C([0,1], B)$ of all continuous $B$-valued functions but I am not sure how to proceed.

• Yes, but if you mean nilpotent in the purely algebraic sense, $x^n=0$ for some $n$, then I don't see how this helps, since the converse fails. If you meant nilpotent in the sense that $||x^n||^{1/n}\to0$ then this gives a solution, but you said "commutative ring"... Found an example before sticking my neck out on algebra: Say $0<a_{j+1}<a_j$ and $(a_1a_2\dots a_n)^{1/n}\to0$. Let $T$ be a weighted left shift on $\ell^2(\mathbb N)$: $Tx=(a_1x_2,a_2x_3,\dots).$ Say $A$ is the Banach algebra generated by $T$. Then $||T^n||^{1/n}\to0$, so $T$ is in every maximal ideal, but $T^n\ne0$. – David C. Ullrich Jul 25 '15 at 2:10
• I meant nilpotent in the algebraic sense. So if $B$ contains a nonzero nilpotent element, you can't embed it in an abelian semi-simple Banach algebra. – Robert Israel Jul 25 '15 at 2:24
Clearly not. Maximal ideals are the kernels of complex homomorphisms $\phi$ with $\phi(e)=1$. Say $A$ is not semisimple. This says there exists $x\in A$ with $x\ne0$ but $\phi(x)=0$ for every complex homomorphism $\phi$ of $A$.
Now suppose $A$ is a subalgebra of $B$, and $\phi$ is a complex homomorphism of $B$ with $\phi(e_B)=1$. Let $\psi=\phi|_A$. Then $\psi$ is a complex homomorphism of $A$, so $\psi(x)=\phi(x)=0$. Hence $B$ is not semisimple.