Unital commutative Banach algebra A, A/radical(A) has no quasinilpotent elements

Question

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?

Answer 1

Assume $x\in A$ such that $x+rad(A)$ is nilpotent in $A/rad(A)$. This implies that for all $\varepsilon > 0$ there is an $n\in \mathbb N$ and $r\in rad(A)$ 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 $x\in rad(A)$.