Lemma 10.95.5. Let $R \to S$ be a ring map, and let $M$ be an $R$-module. Suppose $M \otimes _ R S = \bigoplus _{i \in I} Q_ i$ is a direct sum of countably generated $S$-modules $Q_ i$. If $N$ is a countably generated submodule of $M$, then there is a countably generated submodule $N'$ of $M$ such that $N' \supset N$ and $\mathop{\mathrm{Im}}(N' \otimes _ R S \to M \otimes _ R S) = \bigoplus _{i \in I'} Q_ i$ for some subset $I' \subset I$.
Proof. Let $N'_0 = N$. We construct by induction an increasing sequence of countably generated submodules $N'_{\ell } \subset M$ for $\ell = 0, 1, 2, \ldots $ such that: if $I'_{\ell }$ is the set of $i \in I$ such that the projection of $\mathop{\mathrm{Im}}(N'_{\ell } \otimes _ R S \to M \otimes _ R S)$ onto $Q_ i$ is nonzero, then $\mathop{\mathrm{Im}}(N'_{\ell + 1} \otimes _ R S \to M \otimes _ R S)$ contains $Q_ i$ for all $i \in I'_{\ell }$. To construct $N'_{\ell + 1}$ from $N'_\ell $, let $Q$ be the sum of (the countably many) $Q_ i$ for $i \in I'_{\ell }$, choose $P$ as in Lemma 10.95.4, and then let $N'_{\ell + 1} = N'_{\ell } + P$. Having constructed the $N'_{\ell }$, just take $N' = \bigcup _{\ell } N'_{\ell }$ and $I' = \bigcup _{\ell } I'_{\ell }$. $\square$
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