
Lemma 10.126.4. 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 10.126.3. Thus (1) implies (2).

Consider the category $\mathcal{I}$ of Lemma 10.126.1. 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. $\square$

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