The Stacks project

Proof. Write $M = \mathop{\mathrm{colim}}\nolimits _{i\in I} M_ i$ as a directed colimit of finitely presented $R$-modules $M_ i$. Using Proposition 10.88.6, we see that we have to prove that for each $i \in I$ there exists $i \leq j$, $j\in I$ such that $M_ i\rightarrow M_ j$ dominates $M_ i\rightarrow M$.

Take $N$ the pushout

Then the lemma is equivalent to the existence of $j$ such that $M_ j\rightarrow N$ is universally injective, see Lemma 10.88.4. Observe that the tensorization by $S$

Is a pushout diagram. So because $M \otimes _ R S = \mathop{\mathrm{colim}}\nolimits _{i\in I} M_ i \otimes _ R S$ expresses $M\otimes _ R S$ as a colimit of $S$-modules of finite presentation, and $M\otimes _ R S$ is Mittag-Leffler, there exists $j \geq i$ such that $M_ j\otimes _ R S\rightarrow N\otimes _ R S$ is universally injective. So using that $R\rightarrow S$ is faithfully flat we conclude that $M_ j\rightarrow N$ is universally injective too. $\square$

Proof. Say $M \otimes _ R S$ is generated by the elements $y_ i$, $i = 1, 2, 3, \ldots $. Write $y_ i = \sum _{j = 1, \ldots , n_ i} x_{ij} \otimes s_{ij}$ for some $n_ i \geq 0$, $x_{ij} \in M$ and $s_{ij} \in S$. Denote $M' \subset M$ the submodule generated by the countable collection of elements $x_{ij}$. Then $M' \otimes _ R S \to M \otimes _ R S$ is surjective as the image contains the generators $y_ i$. Since $S$ is faithfully flat over $R$ we conclude that $M' = M$ as desired. $\square$

Proof. Follows from Lemmas 10.83.2, 10.95.1, and 10.95.2 and Theorem 10.93.3. $\square$

Proof. Let $y_1, y_2, \ldots $ be generators for $Q$ and write $y_ j = \sum _ k x_{jk} \otimes s_{jk}$ for some $x_{jk} \in M$ and $s_{jk} \in S$. Then take $P$ be the submodule of $M$ generated by the $x_{jk}$. $\square$

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 $I' = \bigcup _{\ell } I'_{\ell }$. $\square$

Proof. We are going to construct a Kaplansky dévissage of $M$ to show that it is a direct sum of projective modules and hence projective. By Theorem 10.84.5 we can write $M \otimes _ R S = \bigoplus _{i \in I} Q_ i$ as a direct sum of countably generated $S$-modules $Q_ i$. Choose a well-ordering on $M$. Using transfinite recursion we are going to define an increasing family of submodules $M_{\alpha }$ of $M$, one for each ordinal $\alpha $, such that $M_{\alpha } \otimes _ R S$ is a direct sum of some subset of the $Q_ i$.

For $\alpha = 0$ let $M_0 = 0$. If $\alpha $ is a limit ordinal and $M_{\beta }$ has been defined for all $\beta < \alpha $, then define $M_\alpha = \bigcup _{\beta < \alpha } M_{\beta }$. Since each $M_{\beta } \otimes _ R S$ for $\beta < \alpha $ is a direct sum of a subset of the $Q_ i$, the same will be true of $M_{\alpha } \otimes _ R S$. If $\alpha + 1$ is a successor ordinal and $M_{\alpha }$ has been defined, then define $M_{\alpha + 1}$ as follows. If $M_{\alpha } = M$, then let $M_{\alpha +1} = M$. Otherwise choose the smallest $x \in M$ (with respect to the fixed well-ordering) such that $x \notin M_{\alpha }$. Since $S$ is flat over $R$, $(M/M_{\alpha }) \otimes _ R S = M \otimes _ R S/M_{\alpha } \otimes _ R S$, so since $M_{\alpha } \otimes _ R S$ is a direct sum of some $Q_ i$, the same is true of $(M/M_{\alpha }) \otimes _ R S$. By Lemma 10.95.5, we can find a countably generated $R$-submodule $P$ of $M/M_{\alpha }$ containing the image of $x$ in $M/M_{\alpha }$ and such that $P \otimes _ R S$ (which equals $\mathop{\mathrm{Im}}(P \otimes _ R S \to M \otimes _ R S)$ since $S$ is flat over $R$) is a direct sum of some $Q_ i$. Since $M \otimes _ R S = \bigoplus _{i \in I} Q_ i$ is projective and projectivity passes to direct summands, $P \otimes _ R S$ is also projective. Thus by Lemma 10.95.3, $P$ is projective. Finally we define $M_{\alpha + 1}$ to be the preimage of $P$ in $M$, so that $M_{\alpha + 1}/M_{\alpha } = P$ is countably generated and projective. In particular $M_{\alpha }$ is a direct summand of $M_{\alpha + 1}$ since projectivity of $M_{\alpha + 1}/M_{\alpha }$ implies the sequence $0 \to M_{\alpha } \to M_{\alpha + 1} \to M_{\alpha + 1}/M_{\alpha } \to 0$ splits.

Transfinite induction on $M$ (using the fact that we constructed $M_{\alpha + 1}$ to contain the smallest $x \in M$ not contained in $M_{\alpha }$) shows that each $x \in M$ is contained in some $M_{\alpha }$. Thus, there is some large enough ordinal $S$ satisfying: for each $x \in M$ there is $\alpha \in S$ such that $x \in M_{\alpha }$. This means $(M_{\alpha })_{\alpha \in S}$ satisfies property (1) of a Kaplansky dévissage of $M$. The other properties are clear by construction. We conclude $M = \bigoplus _{\alpha + 1 \in S} M_{\alpha + 1}/M_{\alpha }$. Since each $M_{\alpha + 1}/M_{\alpha }$ is projective by construction, $M$ is projective. $\square$