The Stacks project

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. $\square$

Comments (2)

Comment #8086 by Laurent Moret-Bailly on

It may be worth pointing out that is of finite type. Namely, there is of finite type such that , and since $F' is minimal.

There are also:

  • 4 comment(s) on Section 10.89: Interchanging direct products with tensor

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