Lemma 16.3.5. Let $R \to \Lambda $ be a ring map. If $\Lambda $ is a filtered colimit of smooth $R$-algebras, then $\Lambda $ is a filtered colimit of standard smooth $R$-algebras.

**Proof.**
Let $A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. According to Algebra, Lemma 10.127.4 we have to factor this map through a standard smooth algebra, and we know we can factor it as $A \to B \to \Lambda $ with $B$ smooth over $R$. Choose an $R$-algebra map $B \to C$ with a retraction $C \to B$ such that $C$ is standard smooth over $R$, see Lemma 16.3.4. Then the desired factorization is $A \to B \to C \to B \to \Lambda $.
$\square$

## 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Your email address will not be published.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)