The Stacks project

Lemma 61.8.1. Given a ring $A$ there exists a faithfully flat ind-étale $A$-algebra $C$ such that every faithfully flat étale ring map $C \to B$ has a section.

Proof. Set $T^1(A) = T(A)$ and $T^{n + 1}(A) = T(T^ n(A))$. Let

\[ C = \mathop{\mathrm{colim}}\nolimits T^ n(A) \]

This algebra is faithfully flat over each $T^ n(A)$ and in particular over $A$, see Algebra, Lemma 10.39.20. Moreover, $C$ is ind-étale over $A$ by Lemma 61.7.4. If $C \to B$ is étale, then there exists an $n$ and an étale ring map $T^ n(A) \to B'$ such that $B = C \otimes _{T^ n(A)} B'$, see Algebra, Lemma 10.143.3. If $C \to B$ is faithfully flat, then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective, hence $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective. In other words, $T^ n(A) \to B'$ is faithfully flat. By our construction, there is a $T^ n(A)$-algebra map $B' \to T^{n + 1}(A)$. This induces a $C$-algebra map $B \to C$ which finishes the proof. $\square$


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Your email address will not be published.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 097R. Beware of the difference between the letter 'O' and the digit '0'.