The Stacks project

Lemma 15.42.8. Let $A \to B$ be a flat local homomorphism of Noetherian local rings such that $\mathfrak m_ A B = \mathfrak m_ B$ and $\kappa (\mathfrak m_ A) = \kappa (\mathfrak m_ B)$. Then $A \to B$ induces an isomorphism $A^\wedge \to B^\wedge $ of completions.

Proof. By Algebra, Lemma 10.96.7 we see that $B^\wedge $ is the $\mathfrak m_ A$-adic completion of $B$ and that $A^\wedge \to B^\wedge $ is finite. Since $A \to B$ is flat we have $\text{Tor}_1^ A(B, \kappa (\mathfrak m_ A)) = 0$. Hence we see that $B^\wedge $ is flat over $A^\wedge $ by Lemma 15.27.5. Thus $B^\wedge $ is a free $A^\wedge $-module by Algebra, Lemma 10.77.4. Since $A^\wedge \to B^\wedge $ induces an isomorphism $\kappa (\mathfrak m_ A) = A^\wedge /\mathfrak m_ A A^\wedge \to B^\wedge /\mathfrak m_ A B^\wedge = B^\wedge /\mathfrak m_ B B^\wedge = \kappa (\mathfrak m_ B)$ by our assumptions (and Algebra, Lemma 10.95.3), we see that $B^\wedge $ is free of rank $1$. Thus $A^\wedge \to B^\wedge $ is an isomorphism. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AGX. Beware of the difference between the letter 'O' and the digit '0'.