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 $P \to M_ j \to M_{j'}$. $\square$

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