Lemma 10.88.3. Let $f: M \to N$ and $g: M \to M'$ be maps of $R$-modules. Then $g$ dominates $f$ if and only if for any finitely presented $R$-module $Q$, we have $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ Q) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ Q)$.
Proof. Suppose $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ Q) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ Q)$ for all finitely presented modules $Q$. If $Q$ is an arbitrary module, write $Q = \mathop{\mathrm{colim}}\nolimits _{i \in I} Q_ i$ as a colimit of a directed system of finitely presented modules $Q_ i$. Then $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_{Q_ i}) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_{Q_ i})$ for all $i$. Since taking directed colimits is exact and commutes with tensor product, it follows that $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ Q) \subset \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ Q)$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)