Lemma 10.89.6. Let M be a flat Mittag-Leffler module over R. Let F be an R-module and let x \in F \otimes _ R M. Then there exists a smallest submodule F' \subset F such that x \in F' \otimes _ R M. Also, F' is a finite R-module.
Proof. Since M is flat we have F' \otimes _ R M \subset F \otimes _ R M if F' \subset F is a submodule, hence the statement makes sense. Let I = \{ F' \subset F \mid x \in F' \otimes _ R M\} and for i \in I denote F_ i \subset F the corresponding submodule. Then x maps to zero under the map
F \otimes _ R M \longrightarrow \prod (F/F_ i \otimes _ R M)
whence by Proposition 10.89.5 x maps to zero under the map
F \otimes _ R M \longrightarrow \left(\prod F/F_ i\right) \otimes _ R M
Since M is flat the kernel of this arrow is (\bigcap F_ i) \otimes _ R M which proves that F' = \bigcap F_ i. To see that F' is a finite module, suppose that x = \sum _{j = 1, \ldots , m} f_ j \otimes m_ j with f_ j \in F' and m_ j \in M. Then x \in F'' \otimes _ R M where F'' \subset F' is the submodule generated by f_1, \ldots , f_ m. Of course then F'' = F' and we conclude the final statement holds. \square
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Comment #8086 by Laurent Moret-Bailly on
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