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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