Shilov Boundary and Topological Divisors of Zero

Let $E$ be a field complete with respect to a nontrivial Archimedean or non-Archimedean ultrametric absolute value and let $(A,\|\cdot \|)$ be a commutative normed $E$-algebra with unity whose spectral seminorm is $\|\cdot \|_{\mathrm {si}}$. Let $\operatorname {Mult}(A,\|\cdot \|)$ be the set of continuous multiplicative seminorms of $A$ and let $\mathcal S$ be the Shilov boundary for $(A,\|\cdot \|_{\mathrm {si}})$. An element $\psi$ of $\operatorname {Mult}(A,\|\cdot \|_{\mathrm {si}})$ belongs to $\mathcal S$ if and only if, for every neighborhood $\mathcal U$ of $\psi$ in $\operatorname {Mult}(A,\|\cdot \|)$, there exist $\theta\in{\mathcal U}$ and $g\in A$ that satisfy $\|g\|_{\mathrm {si}}=\theta (g)$ and $\gamma (g)<\|g\|_{\mathrm {si}}$ for all $\gamma \in {\mathcal S}\setminus U$. Suppose that $A$ is uniform and $f\in A$. Then, $f$ is a topological divisor of zero if and only if there exists $\psi\in\mathcal S$ such that $\psi(f)=0$. Moreover, if $f$ is not a divisor of zero, then it is a topological divisor of zero if and only if the ideal $fA$ is not closed in $A$. Suppose that $A$ is ultrametric, complete, and Noetherian. All topological divisors of zero are divisors of zero. This applies to affinoid algebras. Let $A$ be a Krasner algebra $H(D)$ without nontrivial idempotents: an element $f\in H(D)$ is a topological divisor of zero if and only if $fH(D)$ is a closed ideal; moreover, $H(D)$ is a principal ideal ring if and only if it has no topological divisors of zero but $0$ (this new condition adds to the well-known set of equivalent conditions found in 1969).