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

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?

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