Lemma 10.95.1. 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 Mittag-Leffler, then $M$ is Mittag-Leffler.

## 10.95 Descending properties of modules

We address the faithfully flat descent of the properties from Theorem 10.93.3 that characterize projectivity. In the presence of flatness, the property of being a Mittag-Leffler module descends:

**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. $\square$

Lemma 10.95.2. 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 countably generated, then $M$ is countably generated.

**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.
$\square$

At this point the faithfully flat descent of countably generated projective modules follows easily.

Lemma 10.95.3. 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 countably generated and projective, then $M$ is countably generated and projective.

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

All that remains is to use dévissage to reduce descent of projectivity in the general case to the countably generated case. First, two simple lemmas.

Lemma 10.95.4. Let $R \to S$ be a ring map, let $M$ be an $R$-module, and let $Q$ be a countably generated $S$-submodule of $M \otimes _ R S$. Then there exists a countably generated $R$-submodule $P$ of $M$ such that $\mathop{\mathrm{Im}}(P \otimes _ R S \to M \otimes _ R S)$ contains $Q$.

**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 $x_{jk}$.
$\square$

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

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. $\square$

## 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #5778 by alexis bouthier on

Comment #5781 by Johan on

Comment #5793 by alexis bouthier on

Comment #8510 by saskia kern on