Lemma 10.168.4 (07RH)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.168: Colimits and maps of finite presentation, II
Lemma 10.168.4 (cite)

numberstags

Lemma 10.168.4. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$ -algebras. Assume

$A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is surjective,
$C_0$ is of finite type over $B_0$ .

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is surjective.

Proof. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$ . Pick $b_ j \in A \otimes _{A_0} B_0$ mapping to $1 \otimes x_ j$ in $A \otimes _{A_0} C_0$ . For some $i \geq 0$ we can find $b_{j, i} \in A_ i \otimes _{A_0} B_0$ mapping to $b_ j$ . After increasing $i$ we may assume that $b_{j, i}$ maps to $1 \otimes x_ j$ in $A_ i \otimes _{A_0} C_0$ for all $j = 1, \ldots , m$ . Then this $i$ works. $\square$

Comments (2)

Comment #8253 by DatPham on March 11, 2023 at 14:46

I think we need to enlarge $i$ once more to ensure that $b_{j,i}$ maps into $1 \otimes x_j$ in $A_i \otimes_{A_0}C_0$ .

Comment #8895 by Stacks project on April 09, 2024 at 14:41

Thanks and fixed here.

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).

Comments (2)

Post a comment