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Let $A$ be a commutative unital Banach Algebra. Suppose for every $a\in A$, $\|a\|=1$, I get a singular element $b_a$. I know that each such $b_a$ is contained in a proper maximal ideal of $A$. Is it possible that all the $b_a$'s together are in aproper maximal ideal?

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    $\begingroup$ How is $b_a$ related to $a$? Nothing you've told us prevents all $b_a$ from being in the same maximal ideal. In fact nothing you've told us prevents all $b_a = 0$. $\endgroup$ – Robert Israel Jun 27 '16 at 11:04
  • $\begingroup$ Basically, for every $a\in A$ I get an element $c_a$ and a scalar $\lambda_a$ such that $\lambda_a\in \sigma(a+c_a)$. So I am treating $b_a= a+c-\lambda_a$ $\endgroup$ – user346635 Jun 27 '16 at 11:16
  • $\begingroup$ So for example you could have $c_a = -a$ and $\lambda_a = 0$? $\endgroup$ – Robert Israel Jun 27 '16 at 11:20
  • $\begingroup$ I calculated and checked this case is possible for me only when $\|a\|=\frac{1}{\|(\lambda_a-a)^{-1}\|}$in my Banach algebra. So I should take that all $b_a$'s are not zero. And we have that $\lambda_a-a$ are invertible $\endgroup$ – user346635 Jun 27 '16 at 11:59
  • $\begingroup$ Rather than adding conditions one at a time, why not tell us the whole problem? $\endgroup$ – Robert Israel Jun 29 '16 at 19:58

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