The Stacks project

Lemma 10.168.3. 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 finite,

  2. $C_0$ is of finite type over $B_0$.

Then there exists an $i \geq 0$ such that the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is finite.

Proof. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Pick monic polynomials $P_ j \in A \otimes _{A_0} B_0[T]$ such that $P_ j(1 \otimes x_ j) = 0$ in $A \otimes _{A_0} C_0$. For some $i \geq 0$ we can find $P_{j, i} \in A_ i \otimes _{A_0} B_0[T]$ mapping to $P_ j$. Since $\otimes $ commutes with colimits we see that $P_{j, i}(1 \otimes x_ j)$ is zero in $A_ i \otimes _{A_0} C_0$ after possibly increasing $i$. Then this $i$ works. $\square$

Comments (0)

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07RG. Beware of the difference between the letter 'O' and the digit '0'.