I am trying to show that for a unital commutative Banach algebra A, A/radical(A) has no quasinilpotent elements (where radical(A)={x\in A: x quasinilpotent}). I know that rad(A) is a closed ideal, and that the quotient map is continuous, so I'd want to do something like this:
Let x not be quasinilpotent, so $lim_{n\rightarrow \infty} ||x^n||^{1/n} =\lambda \neq 0$. Let $\pi(x)=\overline{x}$, where $\pi(x):A\rightarrow A/rad(A)$ is the canonical quotient map. Suppose that $||\overline{x}^n||^{1/n}=0$. Then it's spectrum $\sigma(\overline{x})=0$, so $\overline{x}$ is not invertible in A/rad(A), and thus generates a proper ideal in A/rad(A). So then $\pi^{-1}(\overline{x})$ generates a proper ideal in A containing rad(A).
From here, if rad(A) is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?