## 10.91 Examples and non-examples of Mittag-Leffler modules

We end this section with some examples and non-examples of Mittag-Leffler modules.

Example 10.91.1. Mittag-Leffler modules.

1. Any finitely presented module is Mittag-Leffler. This follows, for instance, from Proposition 10.88.6 (1). In general, it is true that a finitely generated module is Mittag-Leffler if and only it is finitely presented. This follows from Propositions 10.89.2, 10.89.3, and 10.89.5.

2. A free module is Mittag-Leffler since it satisfies condition (1) of Proposition 10.88.6.

3. By the previous example together with Lemma 10.89.10, projective modules are Mittag-Leffler.

We also want to add to our list of examples power series rings over a Noetherian ring $R$. This will be a consequence the following lemma.

Lemma 10.91.2. Let $M$ be a flat $R$-module. The following are equivalent

1. $M$ is Mittag-Leffler, and

2. if $F$ is a finite free $R$-module and $x \in F \otimes _ R M$, then there exists a smallest submodule $F'$ of $F$ such that $x \in F' \otimes _ R M$.

Proof. The implication (1) $\Rightarrow$ (2) is a special case of Lemma 10.89.6. Assume (2). By Theorem 10.81.4 we can write $M$ as the colimit $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ of a directed system $(M_ i, f_{ij})$ of finite free $R$-modules. By Remark 10.88.8, it suffices to show that the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, R))$ is Mittag-Leffler. In other words, fix $i \in I$ and for $j \geq i$ let $Q_ j$ be the image of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ j, R) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$; we must show that the $Q_ j$ stabilize.

Since $M_ i$ is free and finite, we can make the identification $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M_ j) = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j$ for all $j$. Using the fact that the $M_ j$ are free, it follows that for $j \geq i$, $Q_ j$ is the smallest submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ such that $f_{ij} \in Q_ j \otimes _ R M_ j$. Under the identification $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M) = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$, the canonical map $f_ i: M_ i \to M$ is in $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$. By the assumption on $M$, there exists a smallest submodule $Q$ of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ such that $f_ i \in Q \otimes _ R M$. We are going to show that the $Q_ j$ stabilize to $Q$.

For $j \geq i$ we have a commutative diagram

$\xymatrix{ Q_ j \otimes _ R M_ j \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j \ar[d] \\ Q_ j \otimes _ R M \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M. }$

Since $f_{ij} \in Q_ j \otimes _ R M_ j$ maps to $f_ i \in \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M$, it follows that $f_ i \in Q_ j \otimes _ R M$. Hence, by the choice of $Q$, we have $Q \subset Q_ j$ for all $j \geq i$.

Since the $Q_ j$ are decreasing and $Q \subset Q_ j$ for all $j \geq i$, to show that the $Q_ j$ stabilize to $Q$ it suffices to find a $j \geq i$ such that $Q_ j \subset Q$. As an element of

$\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M = \mathop{\mathrm{colim}}\nolimits _{j \in J} (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j),$

$f_ i$ is the colimit of $f_{ij}$ for $j \geq i$, and $f_ i$ also lies in the submodule

$\mathop{\mathrm{colim}}\nolimits _{j \in J} (Q \otimes _ R M_ j) \subset \mathop{\mathrm{colim}}\nolimits _{j \in J} (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R) \otimes _ R M_ j).$

It follows that for some $j \geq i$, $f_{ij}$ lies in $Q \otimes _ R M_ j$. Since $Q_ j$ is the smallest submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, R)$ with $f_{ij} \in Q_ j \otimes _ R M_ j$, we conclude $Q_ j\subset Q$. $\square$

Lemma 10.91.3. Let $R$ be a Noetherian ring and $A$ a set. Then $M = R^ A$ is a flat and Mittag-Leffler $R$-module.

Proof. Combining Lemma 10.90.5 and Proposition 10.90.6 we see that $M$ is flat over $R$. We show that $M$ satisfies the condition of Lemma 10.91.2. Let $F$ be a free finite $R$-module. If $F'$ is any submodule of $F$ then it is finitely presented since $R$ is Noetherian. So by Proposition 10.89.3 we have a commutative diagram

$\xymatrix{ F' \otimes _ R M \ar[r] \ar[d]^{\cong } & F \otimes _ R M \ar[d]^{\cong } \\ (F')^ A \ar[r] & F^ A }$

by which we can identify the map $F' \otimes _ R M \to F \otimes _ R M$ with $(F')^ A \to F^ A$. Hence if $x \in F \otimes _ R M$ corresponds to $(x_\alpha ) \in F^ A$, then the submodule of $F'$ of $F$ generated by the $x_\alpha$ is the smallest submodule of $F$ such that $x \in F' \otimes _ R M$. $\square$

Lemma 10.91.4. Let $R$ be a Noetherian ring and $n$ a positive integer. Then the $R$-module $M = R[[t_1, \ldots , t_ n]]$ is flat and Mittag-Leffler.

Proof. As an $R$-module, we have $M = R^ A$ for a (countable) set $A$. Hence this lemma is a special case of Lemma 10.91.3. $\square$

Example 10.91.5. Non Mittag-Leffler modules.

1. By Example 10.89.1 and Proposition 10.89.5, $\mathbf{Q}$ is not a Mittag-Leffler $\mathbf{Z}$-module.

2. We prove below (Theorem 10.93.3) that for a flat and countably generated module, projectivity is equivalent to being Mittag-Leffler. Thus any flat, countably generated, non-projective module $M$ is an example of a non-Mittag-Leffler module. For such an example, see Remark 10.78.4.

3. Let $k$ be a field. Let $R = k[[x]]$. The $R$-module $M = \prod _{n \in \mathbf{N}} R/(x^ n)$ is not Mittag-Leffler. Namely, consider the element $\xi = (\xi _1, \xi _2, \xi _3, \ldots )$ defined by $\xi _{2^ m} = x^{2^{m - 1}}$ and $\xi _ n = 0$ else, so

$\xi = (0, x, 0, x^2, 0, 0, 0, x^4, 0, 0, 0, 0, 0, 0, 0, x^8, \ldots )$

Then the annihilator of $\xi$ in $M/x^{2^ m}M$ is generated $x^{2^{m - 1}}$ for $m \gg 0$. But if $M$ was Mittag-Leffler, then there would exist a finite $R$-module $Q$ and an element $\xi ' \in Q$ such that the annihilator of $\xi '$ in $Q/x^ l Q$ agrees with the annihilator of $\xi$ in $M/x^ l M$ for all $l \geq 1$, see Proposition 10.88.6 (1). Now you can prove there exists an integer $a \geq 0$ such that the annihilator of $\xi '$ in $Q/x^ l Q$ is generated by either $x^ a$ or $x^{l - a}$ for all $l \gg 0$ (depending on whether $\xi ' \in Q$ is torsion or not). The combination of the above would give for all $l = 2^ m >> 0$ the equality $a = l/2$ or $l - a = l/2$ which is nonsensical.

4. The same argument shows that $(x)$-adic completion of $\bigoplus _{n \in \mathbf{N}} R/(x^ n)$ is not Mittag-Leffler over $R = k[[x]]$ (hint: $\xi$ is actually an element of this completion).

5. Let $R = k[a, b]/(a^2, ab, b^2)$. Let $S$ be the finitely presented $R$-algebra with presentation $S = R[t]/(at - b)$. Then as an $R$-module $S$ is countably generated and indecomposable (details omitted). On the other hand, $R$ is Artinian local, hence complete local, hence a henselian local ring, see Lemma 10.153.9. If $S$ was Mittag-Leffler as an $R$-module, then it would be a direct sum of finite $R$-modules by Lemma 10.153.13. Thus we conclude that $S$ is not Mittag-Leffler as an $R$-module.

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