
Lemma 10.96.10. Let $A$ be a Noetherian ring. Let $I, J \subset A$ be ideals. If $A$ is $I$-adically complete and $A/I$ is $J$-adically complete, then $A$ is $J$-adically complete.

Proof. Let $B$ be the $(I + J)$-adic completion of $A$. By Lemma 10.96.2 $B/IB$ is the $J$-adic completion of $A/I$ hence isomorphic to $A/I$ by assumption. Moreover $B$ is $I$-adically complete by Lemma 10.95.8. Hence $B$ is a finite $A$-module by Lemma 10.95.12. By Nakayama's lemma (Lemma 10.19.1 using $I$ is in the Jacobson radical of $A$ by Lemma 10.95.6) we find that $A \to B$ is surjective. The map $A \to B$ is flat by Lemma 10.96.2. The image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ contains $V(I)$ and as $I$ is contained in the Jacobson radical of $A$ we find $A \to B$ is faitfully flat (Lemma 10.38.16). Thus $A \to B$ is injective. Thus $A$ is complete with respect to $I + J$, hence a fortiori complete with respect to $J$. $\square$

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