Lemma 87.6.5. Let \varphi : A \to B be a continuous map of linearly topologized rings. If \varphi is taut, A is pre-adic and has a finitely generated ideal of definition, and B is complete, then B is adic and has a finitely generated ideal of definition and the ring map \varphi is adic.
Proof. Choose a finitely generated ideal of definition I of A. Let J_ n be the closure of \varphi (I^ n)B in B. Since B is complete we have B = \mathop{\mathrm{lim}}\nolimits B/J_ n. Let B' = \mathop{\mathrm{lim}}\nolimits B/I^ nB be the I-adic completion of B. By Algebra, Lemma 10.96.3, the I-adic topology on B' is complete and B'/I^ nB' = B/I^ nB. Thus the ring map B' \to B is continuous and has dense image as B' \to B/I^ nB \to B/J_ n is surjective for all n. Finally, the map B' \to B is taut because (I^ nB')B = I^ nB and A \to B is taut. By Lemma 87.5.10 we see that B' \to B is open and surjective. Thus the topology on B is the I-adic topology and the proof is complete. \square
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