Topological divisors of zero and Shilov boundary

Let L be a field complete for a non-trivial ultrametric absolute value and let ( A , ‖ ⋅ ‖ ) be a commutative normed L-algebra with unity whose spectral semi-norm is ‖ ⋅ ‖ s i . Let Mult ( A , ‖ ⋅ ‖ ) be the set of continuous multiplicative semi-norms of A, let S be the Shilov boundary for ( A , ‖ ⋅ ‖ s i ) and let ψ ∈ Mult ( A , ‖ ⋅ ‖ s i ) . Then ψ belongs to S if and only if for every neighborhood U of ψ in Mult ( A , ‖ ⋅ ‖ ) , there exists θ ∈ U and g ∈ A satisfying ‖ g ‖ s i = θ ( g ) and γ ( g ) < ‖ g ‖ s i ∀ γ ∈ S ∖ U . Suppose A is uniform, let f ∈ A and let Z ( f ) = { ϕ ∈ Mult ( A , ‖ ⋅ ‖ ) | ϕ ( f ) = 0 } . Then f is a topological divisor of zero if and only if there exists ψ ∈ S such that ψ ( f ) = 0 . Suppose now A is complete. 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 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 non-trivial idempotents: an element f ∈ H ( D ) is a topological divisor of zero if and only if f H ( D ) is not 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).