Lemma 87.5.6. Let A \to B and A \to C be continuous homomorphisms of linearly topologized rings. If A \to B is taut, then C \to B \widehat{\otimes }_ A C is taut.
Proof. Let K \subset C be an open ideal. Choose any open ideal I \subset A whose image in C is contained in J. By assumption the closure J of IB is open. Since A \to B is taut we see that B \widehat{\otimes }_ A C is the limit of the rings B/J \otimes _{A/I} C/K over all choices of K and I, i.e, the ideals J(B \widehat{\otimes }_ A C) + K(B \widehat{\otimes }_ A C) form a fundamental system of open ideals. Now, since B \to B \widehat{\otimes }_ A C is continuous we see that J maps into the closure of K(B \widehat{\otimes }_ A C) (as I maps into K). Hence this closure is equal to J(B \widehat{\otimes }_ A C) + K(B \widehat{\otimes }_ A C) and the proof is complete. \square
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