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 category^{1} $\mathcal{I}$ 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. $\square$

## 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Your email address will not be published.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 (4)

Comment #3253 by Dario Weißmann on

Comment #3349 by Johan on

Comment #4890 by Rankeya on

Comment #5166 by Johan on