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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.

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Isn't a nilpotent element of a commutative ring contained in every maximal ideal?

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  • $\begingroup$ 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 $T^n\ne0$. $\endgroup$ – David C. Ullrich Jul 25 '15 at 2:10
  • $\begingroup$ 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. $\endgroup$ – Robert Israel Jul 25 '15 at 2:24
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EDIT: Robert Israel mentioning the words "commutative ring" got me thinking. My answer was somewhat stupid. Perfectly correct, but stupid not to note that (unless I'm missing something) exactly the same argument works in the context of commutative rings instead of Banach algebras. Take what's below and take "complex homomorphism" as a nonstandard spelling of "homomorphism onto some field"...


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.

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