# Ideal generated by a set of singular elements in a Banach algebra.

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?

• 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 $b_a = 0$. – Robert Israel Jun 27 '16 at 11:04
• 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 $b_a= a+c-\lambda_a$ – user346635 Jun 27 '16 at 11:16
• So for example you could have $c_a = -a$ and $\lambda_a = 0$? – Robert Israel Jun 27 '16 at 11:20
• 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 – user346635 Jun 27 '16 at 11:59
• Rather than adding conditions one at a time, why not tell us the whole problem? – Robert Israel Jun 29 '16 at 19:58