The Stacks project

16.12 The main theorem

In this section we wrap up the discussion.

Proof. By Lemma 16.8.4 it suffices to prove this for $k \to \Lambda $ where $\Lambda $ is Noetherian and geometrically regular over $k$. Let $k \to A \to \Lambda $ be a factorization with $A$ a finite type $k$-algebra. It suffices to construct a factorization $A \to B \to \Lambda $ with $B$ of finite type such that $\mathfrak h_ B = \Lambda $, see Lemma 16.2.8. Hence we may perform Noetherian induction on the ideal $\mathfrak h_ A$. Pick a prime $\mathfrak q \supset \mathfrak h_ A$ such that $\mathfrak q$ is minimal over $\mathfrak h_ A$. It now suffices to resolve $k \to A \to \Lambda \supset \mathfrak q$ (as defined in the text following Situation 16.9.1). If the characteristic of $k$ is zero, this follows from Lemma 16.10.3. If the characteristic of $k$ is $p > 0$, this follows from Lemma 16.11.4. $\square$


Comments (3)

Comment #6666 by 蒙古上单 on

I have a question: (Maybe stupid): Is any formally etale ring maps of Notherian rings a filtered colimt of etale ring maps?

Comment #6670 by on

Let . The map is formally etale, see Lemma 110.43.4. But since the fraction field extension is nonalgebraic, the ring cannot be a filtered colimit of etale -algebras.

Comment #6674 by 蒙古上单 on

Thank you very much!


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Your email address will not be published.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 07GB. Beware of the difference between the letter 'O' and the digit '0'.