Lemma 10.127.1. 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 category1 $\mathcal{I}$ is nonempty as $R \to R$ 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 10.14.2). 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. $\square$

 To avoid set theoretical difficulties we consider only $A' \to A$ such that $A'$ is a quotient of $R[x_1, x_2, x_3, \ldots ]$.

Comment #3253 by Dario Weißmann on

On the set theoretical issue: It think it would be more natural to consider quotients of $R[(x_n)_{n\in \mathbf{N}}]$ (that are finitely presented).

Comment #3349 by on

Yes, that would work as well and maybe is more clear. But your comment is evidence that the footnote did its work of making you think it through if you wanted to think it through.

Comment #4890 by Rankeya on

First sentence of proof should say: The category $\mathcal{I}$ is non-empty as $R \rightarrow R$ is an object of it.

Comment #5166 by on

OK, I fixed this and I also changed the footnote. See here.

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