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 $J$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: