The Stacks project

Theorem 10.95.6. Let $R \to S$ be a faithfully flat ring map. Let $M$ be an $R$-module. If the $S$-module $M \otimes _ R S$ is projective, then $M$ is projective.

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 $M_{\beta } = \bigcup _{\beta < \alpha } M_{\beta }$. 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. $\square$

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