The Stacks project

Lemma 15.12.2. Let $(A, I)$ be a pair. Let $(A^ h, I^ h)$ be as in Lemma 15.12.1. Then $A \to A^ h$ is flat, $I^ h = IA^ h$ and $A/I^ n \to A^ h/I^ nA^ h$ is an isomorphism for all $n$.

Proof. In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of ├ętale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism and we have seen that $I^ h = IA^ h$. As an ├ętale ring map is flat (Algebra, Lemma 10.143.3) we conclude that $A \to A^ h$ is flat by Algebra, Lemma 10.39.3. Since each $A \to B$ is flat we find that the maps $A/I^ n \to B/I^ nB$ are isomorphisms as well (for example by Algebra, Lemma 10.101.3). Taking the colimit we find that $A/I^ n = A^ h/I^ nA^ h$ as desired. $\square$

Comments (1)

Comment #5033 by slogan_bot on

Suggested slogan: "Henselization is a flat map that is an isomorphism mod I^n"

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Your email address will not be published.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 0AGU. Beware of the difference between the letter 'O' and the digit '0'.