Lemma 10.126.1. Let $R \to A$ be a ring map. Consider the category $\mathcal{I}$ of all diagrams of $R$-algebra maps $A' \to A$ with $A'$ finitely presented over $R$. Then $\mathcal{I}$ is filtered, and the colimit of the $A'$ over $\mathcal{I}$ is isomorphic to $A$.

**Proof.**
The category^{1} $\mathcal{I}$ is nonempty as $R \to A$ is an object of it. Consider a pair of objects $A' \to A$, $A'' \to A$ of $\mathcal{I}$. Then $A' \otimes _ R A'' \to A$ is in $\mathcal{I}$ (use Lemmas 10.6.2 and 10.13.2). The ring maps $A' \to A' \otimes _ R A''$ and $A'' \to A' \otimes _ R A''$ define arrows in $\mathcal{I}$ thereby proving the second defining property of a filtered category, see Categories, Definition 4.19.1. Finally, suppose that we have two morphisms $\sigma , \tau : A' \to A''$ in $\mathcal{I}$. If $x_1, \ldots , x_ r \in A'$ are generators of $A'$ as an $R$-algebra, then we can consider $A''' = A''/(\sigma (x_ i) - \tau (x_ i))$. This is a finitely presented $R$-algebra and the given $R$-algebra map $A'' \to A$ factors through the surjection $\nu : A'' \to A'''$. Thus $\nu $ is a morphism in $\mathcal{I}$ equalizing $\sigma $ and $\tau $ as desired.

The fact that our index category is cofiltered means that we may compute the value of $B = \mathop{\mathrm{colim}}\nolimits _{A' \to A} A'$ in the category of sets (some details omitted; compare with the discussion in Categories, Section 4.19). To see that $B \to A$ is surjective, for every $a \in A$ we can use $R[x] \to A$, $x \mapsto a$ to see that $a$ is in the image of $B \to A$. Conversely, if $b \in B$ is mapped to zero in $A$, then we can find $A' \to A$ in $\mathcal{I}$ and $a' \in A'$ which maps to $b$. Then $A'/(a') \to A$ is in $\mathcal{I}$ as well and the map $A' \to B$ factors as $A' \to A'/(a') \to B$ which shows that $b = 0$ as desired. $\square$

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## Comments (2)

Comment #3253 by Dario Weißmann on

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