The Stacks project

Proposition 10.89.5. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is Mittag-Leffler.

  2. For every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map $M \otimes _ R \left( \prod _{\alpha } Q_{\alpha } \right) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$ is injective.

Proof. First we prove (1) implies (2). Suppose $M$ is Mittag-Leffler and let $x$ be in the kernel of $M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$. Write $M$ as a colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system of finitely presented modules $M_ i$. Then $M \otimes _ R (\prod _{\alpha } Q_{\alpha })$ is the colimit of $M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha })$. So $x$ is the image of an element $x_ i \in M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha })$. We must show that $x_ i$ maps to $0$ in $M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha })$ for some $j \geq i$. Since $M$ is Mittag-Leffler, we may choose $j \geq i$ such that $M_ i \to M_ j$ and $M_ i \to M$ dominate each other. Then consider the commutative diagram

\[ \xymatrix{ M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \\ M_ i \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[d] \ar[u] & \prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \ar[d] \ar[u] \\ M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } & \prod _{\alpha } (M_ j \otimes _ R Q_{\alpha }) } \]

whose bottom two horizontal maps are isomorphisms, according to Proposition 10.89.3. Since $x_ i$ maps to $0$ in $\prod _{\alpha } (M \otimes _ R Q_{\alpha })$, its image in $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha })$ is in the kernel of the map $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$. But this kernel equals the kernel of $\prod _{\alpha } (M_ i \otimes _ R Q_{\alpha }) \to \prod _{\alpha } (M_ j \otimes _ R Q_{\alpha })$ according to the choice of $j$. Thus $x_ i$ maps to $0$ in $\prod _{\alpha } (M_ j \otimes _ R Q_{\alpha })$ and hence to $0$ in $M_ j \otimes _ R (\prod _{\alpha } Q_{\alpha })$.

Now suppose (2) holds. We prove $M$ satisfies formulation (1) of being Mittag-Leffler from Proposition 10.88.6. Let $f: P \to M$ be a map from a finitely presented module $P$ to $M$. Choose a set $B$ of representatives of the isomorphism classes of finitely presented $R$-modules. Let $A$ be the set of pairs $(Q, x)$ where $Q \in B$ and $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to M \otimes Q)$. For $\alpha = (Q, x) \in A$, we write $Q_{\alpha }$ for $Q$ and $x_{\alpha }$ for $x$. Consider the commutative diagram

\[ \xymatrix{ M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \\ P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[u] & \prod _{\alpha } (P \otimes _ R Q_{\alpha }) \ar[u] . } \]

The top arrow is an injection by assumption, and the bottom arrow is an isomorphism by Proposition 10.89.3. Let $x \in P \otimes _ R (\prod _{\alpha } Q_{\alpha })$ be the element corresponding to $(x_{\alpha }) \in \prod _{\alpha } (P \otimes _ R Q_{\alpha })$ under this isomorphism. Then $x \in \mathop{\mathrm{Ker}}( P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to M \otimes _ R (\prod _{\alpha } Q_{\alpha }))$ since the top arrow in the diagram is injective. By Lemma 10.89.4, we get a finitely presented module $P'$ and a map $f': P \to P'$ such that $f: P \to M$ factors through $f'$ and $x \in \mathop{\mathrm{Ker}}(P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to P' \otimes _ R (\prod _{\alpha } Q_{\alpha }))$. We have a commutative diagram

\[ \xymatrix{ P' \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } & \prod _{\alpha } (P' \otimes _ R Q_{\alpha }) \\ P \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r]^{\cong } \ar[u] & \prod _{\alpha } (P \otimes _ R Q_{\alpha }) \ar[u] . } \]

where both the top and bottom arrows are isomorphisms by Proposition 10.89.3. Thus since $x$ is in the kernel of the left vertical map, $(x_{\alpha })$ is in the kernel of the right vertical map. This means $x_{\alpha } \in \mathop{\mathrm{Ker}}(P \otimes _ R Q_{\alpha } \to P' \otimes _ R Q_{\alpha })$ for every $\alpha \in A$. By the definition of $A$ this means $\mathop{\mathrm{Ker}}(P \otimes _ R Q \to P' \otimes _ R Q) \supset \mathop{\mathrm{Ker}}(P \otimes _ R Q \to M \otimes _ R Q)$ for all finitely presented $Q$ and, since $f: P \to M$ factors through $f': P \to P'$, actually equality holds. By Lemma 10.88.3, $f$ and $f'$ dominate each other. $\square$

Comments (0)

There are also:

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

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 059M. Beware of the difference between the letter 'O' and the digit '0'.