Lemma 87.5.5. Let A \to B and B \to C be continuous homomorphisms of linearly topologized rings. If A \to C is taut, then B \to C is taut.
Proof. Let J \subset B be an open ideal with inverse image I \subset A. Then the closure of JC contains the closure of IC. Hence this closure is open as A \to C is taut. Let I_\lambda be a fundamental system of open ideals of A. Let K_\lambda be the closure of I_\lambda C. Since A \to C is taut, these form a fundamental system of open ideals of C. Denote J_\lambda \subset B the inverse image of K_\lambda . Then the closure of J_\lambda C is K_\lambda . Hence we see that the closures of the ideals JC, where J runs over the open ideals of B form a fundamental system of open ideals of C. \square
Comments (0)