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

1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is surjective,

2. $C_0$ 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 $b_ j$. Then this $i$ works. $\square$

Comment #8253 by DatPham on

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

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).