The Stacks project

Lemma 15.7.6. Let $A, A', B, B', C, C', D, D', I, M', M, N, \varphi $ be as in Lemma 15.7.2. If

  1. $N$ is finitely presented over $D$ and flat over $B$,

  2. $M'$ finitely presented over $C'$ and flat over $A'$, and

  3. the ring map $B' \to D'$ factors as $B' \to D'' \to D'$ with $B' \to D''$ flat and $D'' \to D'$ of finite presentation,

then $N' = N \times _ M M'$ is finitely presented over $D'$.

Proof. Choose a surjection $D''' = D''[x_1, \ldots , x_ n] \to D'$ with finitely generated kernel $J$. By Algebra, Lemma 10.35.23 it suffices to show that $N'$ is finitely presented as a $D'''$-module. Moreover, $D''' \otimes _{B'} B \to D' \otimes _{B'} B = D$ and $D''' \otimes _{B'} A' \to D' \otimes _{B'} A' = C'$ are surjections whose kernels are generated by the image of $J$, hence $N$ is a finitely presented $D''' \otimes _{B'} B$-module and $M'$ is a finitely presented $D''' \otimes _{B'} A'$-module by Algebra, Lemma 10.35.23 again. Thus we may replace $D'$ by $D'''$ and $D$ by $D''' \otimes _{B'} B$, etc. Since $D'''$ is flat over $B'$, it follows that we may assume that $B' \to D'$ is flat.

Assume $B' \to D'$ is flat. By Lemma 15.7.4 the module $N'$ is finite over $D'$. Choose a surjection $(D')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $D^{\oplus n} \to N$, $(C')^{\oplus n} \to M'$, and $C^{\oplus n} \to M$ with kernels $K_ D$, $K_{C'}$, and $K_ C$. There is a canonical map

\[ K' \longrightarrow K_ D \times _{K_ C} K_{C'} \]

On the other hand, since $N' = N \times _ M M'$ and $D' = D \times _ C C'$ (by Lemma 15.6.8; applied to the flat $B'$-module $D'$) there is also a canonical map $K_ D \times _{K_ C} K_{C'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 10.5.3 the modules $K_ D$ and $K_{C'}$ are finite. We conclude from Lemma 15.7.4 that $K'$ is a finite $D'$-module provided that $K_ D \to K_ C$ and $K_{C'} \to K_ C$ induce isomorphisms $K_ D \otimes _ B A = K_ C = K_{C'} \otimes _{A'} A$. This is true because the flatness assumptions implies the sequences

\[ 0 \to K_ D \to D^{\oplus n} \to N \to 0 \quad \text{and}\quad 0 \to K_{C'} \to (C')^{\oplus n} \to M' \to 0 \]

stay exact upon tensoring, see Algebra, Lemma 10.38.12. $\square$


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Your email address will not be published.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 08KP. Beware of the difference between the letter 'O' and the digit '0'.