The Stacks project

Lemma 10.127.4. Let $R \to \Lambda $ be a ring map. Let $\mathcal{E}$ be a set of $R$-algebras such that each $A \in \mathcal{E}$ is of finite presentation over $R$. Then the following two statements are equivalent

  1. $\Lambda $ is a filtered colimit of elements of $\mathcal{E}$, and

  2. for any $R$ algebra map $A \to \Lambda $ with $A$ of finite presentation over $R$ we can find a factorization $A \to B \to \Lambda $ with $B \in \mathcal{E}$.

Proof. Suppose that $\mathcal{I} \to \mathcal{E}$, $i \mapsto A_ i$ is a filtered diagram such that $\Lambda = \mathop{\mathrm{colim}}\nolimits _ i A_ i$. Let $A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. Then we get a factorization $A \to A_ i \to \Lambda $ by applying Lemma 10.127.3. Thus (1) implies (2).

Consider the category $\mathcal{I}$ of Lemma 10.127.1. By Categories, Lemma 4.19.3 the full subcategory $\mathcal{J}$ consisting of those $A \to \Lambda $ with $A \in \mathcal{E}$ is cofinal in $\mathcal{I}$ and is a filtered category. Then $\Lambda $ is also the colimit over $\mathcal{J}$ by Categories, Lemma 4.17.2. $\square$

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Your email address will not be published.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 07C3. Beware of the difference between the letter 'O' and the digit '0'.