Topological Divisors of Zero and Shilov Boundary
View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Topology and its Applications 153 (2006) 1152–1163 www.elsevier.com/locate/topol Topological divisors of zero and Shilov boundary Alain Escassut Laboratoire de Mathématiques UMR 6620, Université Blaise Pascal, Clermont-Ferrand, Les Cézeaux, 63177 Aubiere cedex, France Received 17 February 2005; accepted 1 March 2005 Abstract Let L be a field complete for a non-trivial ultrametric absolute value and let (A, ·) be a com- mutative normed L-algebra with unity whose spectral semi-norm is ·si.LetMult(A, ·) be the set of continuous multiplicative semi-norms of A,letS be the Shilov boundary for (A, ·si) and let ψ ∈ Mult(A, ·si).Thenψ belongs to S if and only if for every neighborhood U of ψ in Mult(A, ·), there exists θ ∈ U and g ∈ A satisfying gsi = θ(g) and γ(g)<gsi ∀γ ∈ S \ U. Suppose A is uniform, let f ∈ A and let Z(f ) ={φ ∈ Mult(A, ·) | φ(f)= 0}.Thenf is a topo- logical 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 fAis 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 fH(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).
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