The Stacks project

10.89 Interchanging direct products with tensor

Let $M$ be an $R$-module and let $(Q_{\alpha })_{\alpha \in A}$ be a family of $R$-modules. Then there is a canonical map $M \otimes _ R \left( \prod _{\alpha \in A} Q_{\alpha } \right) \to \prod _{\alpha \in A} ( M \otimes _ R Q_{\alpha })$ given on pure tensors by $x \otimes (q_{\alpha }) \mapsto (x \otimes q_{\alpha })$. This map is not necessarily injective or surjective, as the following example shows.

Example 10.89.1. Take $R = \mathbf{Z}$, $M = \mathbf{Q}$, and consider the family $Q_ n = \mathbf{Z}/n$ for $n \geq 1$. Then $\prod _ n (M \otimes Q_ n) = 0$. However there is an injection $\mathbf{Q} \to M \otimes (\prod _ n Q_ n)$ obtained by tensoring the injection $\mathbf{Z} \to \prod _ n Q_ n$ by $M$, so $M \otimes (\prod _ n Q_ n)$ is nonzero. Thus $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n)$ is not injective.

On the other hand, take again $R = \mathbf{Z}$, $M = \mathbf{Q}$, and let $Q_ n = \mathbf{Z}$ for $n \geq 1$. The image of $M \otimes (\prod _ n Q_ n) \to \prod _ n (M \otimes Q_ n) = \prod _ n M$ consists precisely of sequences of the form $(a_ n/m)_{n \geq 1}$ with $a_ n \in \mathbf{Z}$ and $m$ some nonzero integer. Hence the map is not surjective.

We determine below the precise conditions needed on $M$ for the map $M \otimes _ R \left( \prod _{\alpha } Q_{\alpha } \right) \to \prod _{\alpha } (M \otimes _ R Q_{\alpha })$ to be surjective, bijective, or injective for all choices of $(Q_{\alpha })_{\alpha \in A}$. This is relevant because the modules for which it is injective turn out to be exactly Mittag-Leffler modules (Proposition 10.89.5). In what follows, if $M$ is an $R$-module and $A$ a set, we write $M^ A$ for the product $\prod _{\alpha \in A} M$.

Proposition 10.89.2. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is finitely generated.

  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 surjective.

  3. For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes _ R Q^{A} \to (M \otimes _ R Q)^{A}$ is surjective.

  4. For every set $A$, the canonical map $M \otimes _ R R^{A} \to M^{A}$ is surjective.

Proof. First we prove (1) implies (2). Choose a surjection $R^ n \to M$ and consider the commutative diagram

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

The top arrow is an isomorphism and the vertical arrows are surjections. We conclude that the bottom arrow is a surjection.

Obviously (2) implies (3) implies (4), so it remains to prove (4) implies (1). In fact for (1) to hold it suffices that the element $d = (x)_{x \in M}$ of $M^ M$ is in the image of the map $f: M \otimes _ R R^{M} \to M^ M$. In this case $d = \sum _{i = 1}^{n} f(x_ i \otimes a_ i)$ for some $x_ i \in M$ and $a_ i \in R^ M$. If for $x \in M$ we write $p_ x: M^ M \to M$ for the projection onto the $x$-th factor, then

\[ x = p_ x(d) = \sum \nolimits _{i = 1}^{n} p_ x(f(x_ i \otimes a_ i)) = \sum \nolimits _{i=1}^{n} p_ x(a_ i) x_ i. \]

Thus $x_1, \ldots , x_ n$ generate $M$. $\square$

Proposition 10.89.3. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is finitely presented.

  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 bijective.

  3. For every $R$-module $Q$ and every set $A$, the canonical map $M \otimes _ R Q^{A} \to (M \otimes _ R Q)^{A}$ is bijective.

  4. For every set $A$, the canonical map $M \otimes _ R R^{A} \to M^{A}$ is bijective.

Proof. First we prove (1) implies (2). Choose a presentation $R^ m \to R^ n \to M$ and consider the commutative diagram

\[ \xymatrix{ R^ m \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & R^ n \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d]^{\cong } & M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ \prod _{\alpha } (R^ m \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (R^ n \otimes _ R Q_{\alpha }) \ar[r] & \prod _{\alpha } (M \otimes _ R Q_{\alpha }) \ar[r] & 0. } \]

The first two vertical arrows are isomorphisms and the rows are exact. This implies that the map $M \otimes _ R (\prod _{\alpha } Q_{\alpha }) \to \prod _{\alpha } ( M \otimes _ R Q_{\alpha })$ is surjective and, by a diagram chase, also injective. Hence (2) holds.

Obviously (2) implies (3) implies (4), so it remains to prove (4) implies (1). From Proposition 10.89.2, if (4) holds we already know that $M$ is finitely generated. So we can choose a surjection $F \to M$ where $F$ is free and finite. Let $K$ be the kernel. We must show $K$ is finitely generated. For any set $A$, we have a commutative diagram

\[ \xymatrix{ & K \otimes _ R R^ A \ar[r] \ar[d]_{f_3} & F \otimes _ R R^ A \ar[r] \ar[d]_{f_2}^{\cong } & M \otimes _ R R^ A \ar[r] \ar[d]_{f_1}^{\cong } & 0 \\ 0 \ar[r] & K^ A \ar[r] & F^ A \ar[r] & M^ A \ar[r] & 0 . } \]

The map $f_1$ is an isomorphism by assumption, the map $f_2$ is a isomorphism since $F$ is free and finite, and the rows are exact. A diagram chase shows that $f_3$ is surjective, hence by Proposition 10.89.2 we get that $K$ is finitely generated. $\square$

We need the following lemma for the next proposition.

Lemma 10.89.4. Let $M$ be an $R$-module, $P$ a finitely presented $R$-module, and $f: P \to M$ a map. Let $Q$ be an $R$-module and suppose $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to M \otimes Q)$. Then there exists a finitely presented $R$-module $P'$ and a map $f': P \to P'$ such that $f$ factors through $f'$ and $x \in \mathop{\mathrm{Ker}}(P \otimes Q \to P' \otimes Q)$.

Proof. Write $M$ as a colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system of finitely presented modules $M_ i$. Since $P$ is finitely presented, the map $f: P \to M$ factors through $M_ j \to M$ for some $j \in I$. Upon tensoring by $Q$ we have a commutative diagram

\[ \xymatrix{ & M_ j \otimes Q \ar[dr] & \\ P \otimes Q \ar[ur] \ar[rr] & & M \otimes Q . } \]

The image $y$ of $x$ in $M_ j \otimes Q$ is in the kernel of $M_ j \otimes Q \to M \otimes Q$. Since $M \otimes Q = \mathop{\mathrm{colim}}\nolimits _{i \in I} (M_ i \otimes Q)$, this means $y$ maps to $0$ in $M_{j'} \otimes Q$ for some $j' \geq j$. Thus we may take $P' = M_{j'}$ and $f'$ to be the composite $P \to M_ j \to M_{j'}$. $\square$

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. $\square$

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. $\square$

Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:

  1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.

  2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.

Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 } \]

with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$

Lemma 10.89.8. Let $M_1 \to M_2 \to M_3 \to 0$ be an exact sequence of $R$-modules. If $M_1$ is finitely generated and $M_2$ is Mittag-Leffler, then $M_3$ is Mittag-Leffler.

Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, since tensor product is right exact, we have a commutative diagram

\[ \xymatrix{ M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 } \]

with exact rows. By Proposition 10.89.2 the left vertical arrow is surjective. By Proposition 10.89.5 the middle vertical arrow is injective. A diagram chase shows the right vertical arrow is injective. Hence $M_3$ is Mittag-Leffler by Proposition 10.89.5. $\square$

Lemma 10.89.9. If $M = \mathop{\mathrm{colim}}\nolimits M_ i$ is the colimit of a directed system of Mittag-Leffler $R$-modules $M_ i$ with universally injective transition maps, then $M$ is Mittag-Leffler.

Proof. Let $(Q_{\alpha })_{\alpha \in A}$ be a family of $R$-modules. We have to show that $M \otimes _ R (\prod Q_\alpha ) \to \prod M \otimes _ R Q_\alpha $ is injective and we know that $M_ i \otimes _ R (\prod Q_\alpha ) \to \prod M_ i \otimes _ R Q_\alpha $ is injective for each $i$, see Proposition 10.89.5. Since $\otimes $ commutes with filtered colimits, it suffices to show that $\prod M_ i \otimes _ R Q_\alpha \to \prod M \otimes _ R Q_\alpha $ is injective. This is clear as each of the maps $M_ i \otimes _ R Q_\alpha \to M \otimes _ R Q_\alpha $ is injective by our assumption that the transition maps are universally injective. $\square$

Lemma 10.89.10. If $M = \bigoplus _{i \in I} M_ i$ is a direct sum of $R$-modules, then $M$ is Mittag-Leffler if and only if each $M_ i$ is Mittag-Leffler.

Proof. The “only if” direction follows from Lemma 10.89.7 (1) and the fact that a split short exact sequence is universally exact. The converse follows from Lemma 10.89.9 but we can also argue it directly as follows. First note that if $I$ is finite then this follows from Lemma 10.89.7 (2). For general $I$, if all $M_ i$ are Mittag-Leffler then we prove the same of $M$ by verifying condition (1) of Proposition 10.88.6. Let $f: P \to M$ be a map from a finitely presented module $P$. Then $f$ factors as $P \xrightarrow {f'} \bigoplus _{i' \in I'} M_{i'} \hookrightarrow \bigoplus _{i \in I} M_ i$ for some finite subset $I'$ of $I$. By the finite case $\bigoplus _{i' \in I'} M_{i'}$ is Mittag-Leffler and hence there exists a finitely presented module $Q$ and a map $g: P \to Q$ such that $g$ and $f'$ dominate each other. Then also $g$ and $f$ dominate each other. $\square$

Lemma 10.89.11. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $S$ is Mittag-Leffler as an $R$-module, and $M$ is flat and Mittag-Leffler as an $S$-module, then $M$ is Mittag-Leffler as an $R$-module.

Proof. We deduce this from the characterization of Proposition 10.89.5. Namely, suppose that $Q_\alpha $ is a family of $R$-modules. Consider the composition

\[ \xymatrix{ M \otimes _ R \prod _\alpha Q_\alpha = M \otimes _ S S \otimes _ R \prod _\alpha Q_\alpha \ar[d] \\ M \otimes _ S \prod _\alpha (S \otimes _ R Q_\alpha ) \ar[d] \\ \prod _\alpha (M \otimes _ S S \otimes _ R Q_\alpha ) = \prod _\alpha (M \otimes _ R Q_\alpha ) } \]

The first arrow is injective as $M$ is flat over $S$ and $S$ is Mittag-Leffler over $R$ and the second arrow is injective as $M$ is Mittag-Leffler over $S$. Hence $M$ is Mittag-Leffler over $R$. $\square$

Comments (4)

Comment #6214 by Adi Caplan-Bricker on

In the last math typeset line of lemma 05CT, needs an inserted between and . In the next linw, "first arrows" should be "first arrow"

Comment #7140 by Ando on

There is a typographical error in the proof of Proposition 059K; the middle of the upper column of the first commutative diagram should be , not .

Comment #7293 by on

Thanks and fixed here. In the future please leave comments on the page of the tag you are commenting on.

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 059H. Beware of the difference between the letter 'O' and the digit '0'.



View Section 10.89 as pdf