I am trying to show that for a unital commutative Banach algebra $A$, $A/\operatorname{radical}(A)$ has no quasinilpotent elements (where $\operatorname{rad}(A)=\{x\in A: x \,\text{quasinilpotent}\}$). I know that $\operatorname{rad}(A)$ 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 $\pi(x):A\rightarrow A/\operatorname{rad}(A)$ 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 $A/\operatorname{rad}(A)$, and thus generates a proper ideal in $A/\operatorname{rad}(A)$. So then $\pi^{-1}(\overline{x})$ generates a proper ideal in A containing $\operatorname{rad}(A)$.
From here, if $\operatorname{rad}(A)$ 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?
Best Answer
Assume $x\in A$ such that $x+\operatorname{rad}(A)$ is nilpotent in $A \operatorname{rad}(A)$. This implies that for all $\varepsilon > 0$ there is an $n\in \mathbb N$ and $r\in \operatorname{rad}(A)$ for which $|x^n-r|\leq\varepsilon^n$.
$r$ is quasi-nilpotent so there exists $M\in\mathbb N$ s.t. for all $m>M$, $|r^m|\leq\varepsilon^{nm}$
We then show by induction that there exists $A>0$ such that $|x^{nm}|\leq A2^m\varepsilon^{nm}$ for all $m$. We choose $A\geq 1$ such that this is true for all $m\in[0,M]$.
Then if $m>M$, we have $$x^{nm}-r^m=(x^n-r)(x^{n(m-1)}+\ldots+r^{m-1}),$$ so $$|x^{nm}|\leq \varepsilon^n(A2^{m-1}\varepsilon^{n(m-1)}+\ldots+A\varepsilon^{n(m-1)}) + \varepsilon^{nm}\leq A\varepsilon^{nm}(1+1+\ldots 2^{m-1})=A 2^m\varepsilon^{nm}.$$
So we can conclude that $\forall \varepsilon >0 \: \liminf |x^n|^{1/n} < 2\varepsilon $, which means that $x\in \operatorname{rad}(A)$.