The Stacks project

Lemma 10.89.4. Let $M$ be an $R$-module, $P$ a finitely presented $R$-module, and $f: P \to M$ a map. Let $Q$ be an $R$-module and suppose $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to M \otimes Q)$. Then there exists a finitely presented $R$-module $P'$ and a map $f': P \to P'$ such that $f$ factors through $f'$ and $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to P' \otimes Q)$.

Proof. Write $M$ as a colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system of finitely presented modules $M_ i$. Since $P$ is finitely presented, the map $f: P \to M$ factors through $M_ j \to M$ for some $j \in I$. Upon tensoring by $Q$ we have a commutative diagram

\[ \xymatrix{ & M_ j \otimes Q \ar[dr] & \\ P \otimes Q \ar[ur] \ar[rr] & & M \otimes Q . } \]

The image $y$ of $x$ in $M_ j \otimes Q$ is in the kernel of $M_ j \otimes Q \to M \otimes Q$. Since $M \otimes Q = \mathop{\mathrm{colim}}\nolimits _{i \in I} (M_ i \otimes Q)$, this means $y$ maps to $0$ in $M_{j'} \otimes Q$ for some $j' \geq j$. Thus we may take $P' = M_{j'}$ and $f'$ to be the composite $P \to M_ j \to M_{j'}$. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 10.89: Interchanging direct products with tensor

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Your email address will not be published.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 059L. Beware of the difference between the letter 'O' and the digit '0'.