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?

  • $\begingroup$ How is the norm defined on the quotient? $\endgroup$ – Berci Jul 15 '19 at 20:06
  • 1
    $\begingroup$ @Berci The norm in the quotient equals the distance to the ideal. $\endgroup$ – Aweygan Jul 15 '19 at 21:21
  • $\begingroup$ I believe the norm of the quotient is defined inf{||x+rad(A)||} where inf is taken over elements in rad(A) $\endgroup$ – Brendan Mallery Jul 15 '19 at 22:16
  • $\begingroup$ @Brendan Mallery, do you agree with my answer? $\endgroup$ – FXV Jul 21 '19 at 20:40
  • $\begingroup$ Yes, thank you so much! Relatively new to the site, am I supposed to mark this question as answered or something? $\endgroup$ – Brendan Mallery Jul 22 '19 at 17:17

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 $x\in rad(A)$.


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