Lemma 10.91.1. Let $M$ be an $R$-module. Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ where $(M_ i, f_{ij})$ is a directed system of finitely presented $R$-modules. If $M$ is Mittag-Leffler and countably generated, then there is a directed countable subset $I' \subset I$ such that $M \cong \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i$.

## 10.91 Countably generated Mittag-Leffler modules

It turns out that countably generated Mittag-Leffler modules have a particularly simple structure.

**Proof.**
Let $x_1, x_2, \ldots $ be a countable set of generators for $M$. For each $x_ n$ choose $i \in I$ such that $x_ n$ is in the image of the canonical map $f_ i: M_ i \to M$; let $I'_{0} \subset I$ be the set of all these $i$. Now since $M$ is Mittag-Leffler, for each $i \in I'_{0}$ we can choose $j \in I$ such that $j \geq i$ and $f_{ij}: M_ i \to M_ j$ factors through $f_{ik}: M_ i \to M_ k$ for all $k \geq i$ (condition (3) of Proposition 10.87.6); let $I'_1$ be the union of $I'_0$ with all of these $j$. Since $I'_1$ is a countable, we can enlarge it to a countable directed set $I'_{2} \subset I$. Now we can apply the same procedure to $I'_{2}$ as we did to $I'_{0}$ to get a new countable set $I'_{3} \subset I$. Then we enlarge $I'_{3}$ to a countable directed set $I'_{4}$. Continuing in this way—adding in a $j$ as in Proposition 10.87.6 (3) for each $ i \in I'_{\ell }$ if $\ell $ is odd and enlarging $I'_{\ell }$ to a directed set if $\ell $ is even—we get a sequence of subsets $I'_{\ell } \subset I$ for $\ell \geq 0$. The union $I' = \bigcup I'_{\ell }$ satisfies:

$I'$ is countable and directed;

each $x_ n$ is in the image of $f_ i: M_ i \to M$ for some $i \in I'$;

if $i \in I'$, then there is $j \in I'$ such that $j \geq i$ and $f_{ij}: M_ i \to M_ j$ factors through $f_{ik}: M_ i \to M_ k$ for all $k \in I$ with $k \geq i$. In particular $\mathop{\mathrm{Ker}}(f_{ik}) \subset \mathop{\mathrm{Ker}}(f_{ij})$ for $k \geq i$.

We claim that the canonical map $\mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i \to \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i = M$ is an isomorphism. By (2) it is surjective. For injectivity, suppose $x \in \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i$ maps to $0$ in $\mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$. Representing $x$ by an element $\tilde{x} \in M_ i$ for some $i \in I'$, this means that $f_{ik}(\tilde{x}) = 0$ for some $k \in I, k \geq i$. But then by (3) there is $j \in I', j \geq i,$ such that $f_{ij}(\tilde{x}) = 0$. Hence $x = 0$ in $\mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i$. $\square$

Lemma 10.91.1 implies that a countably generated Mittag-Leffler module $M$ over $R$ is the colimit of a system

with each $M_ n$ a finitely presented $R$-module. To see this argue as in the proof of Lemma 10.85.3 to see that a countable directed set has a cofinal subset isomorphic to $(\mathbf{N}, \geq )$. Suppose $R = k[x_1, x_2, x_3, \ldots ]$ and $M = R/(x_ i)$. Then $M$ is finitely generated but not finitely presented, hence not Mittag-Leffler (see Example 10.90.1 part (1)). But of course you can write $M = \mathop{\mathrm{colim}}\nolimits _ n M_ n$ by taking $M_ n = R/(x_1, \ldots , x_ n)$, hence the condition that you can write $M$ as such a limit does not imply that $M$ is Mittag-Leffler.

Lemma 10.91.2. Let $R$ be a ring. Let $M$ be an $R$-module. Assume $M$ is Mittag-Leffler and countably generated. For any $R$-module map $f : P \to M$ with $P$ finitely generated there exists an endomorphism $\alpha : M \to M$ such that

$\alpha : M \to M$ factors through a finitely presented $R$-module, and

$\alpha \circ f = f$.

**Proof.**
Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ as a directed colimit of finitely presented $R$-modules with $I$ countable, see Lemma 10.91.1. The transition maps are denoted $f_{ij}$ and we use $f_ i : M_ i \to M$ to denote the canonical maps into $M$. Set $N = \prod _{s \in I} M_ s$. Denote

so that $(M_ i^*)$ is an inverse system of $R$-modules over $I$. Note that $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) = \mathop{\mathrm{lim}}\nolimits M_ i^*$. As $M$ is Mittag-Leffler, we find for every $i \in I$ an index $k(i) \geq i$ such that

Choose and fix $j \in I$ such that $\mathop{\mathrm{Im}}(P \to M) \subset \mathop{\mathrm{Im}}(M_ j \to M)$. This is possible as $P$ is finitely generated. Set $k = k(j)$. Let $x = (0, \ldots , 0, \text{id}_{M_ k}, 0, \ldots , 0) \in M_ k^*$ and note that this maps to $y = (0, \ldots , 0, f_{jk}, 0, \ldots , 0) \in M_ j^*$. By our choice of $k$ we see that $y \in E_ j$. By Example 10.85.2 the transition maps $E_ i \to E_ j$ are surjective for each $i \geq j$ and $\mathop{\mathrm{lim}}\nolimits E_ i = \mathop{\mathrm{lim}}\nolimits M_ i^* = \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$. Hence Lemma 10.85.3 guarantees there exists an element $z \in \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ which maps to $y$ in $E_ j \subset M_ j^*$. Let $z_ k$ be the $k$th component of $z$. Then $z_ k : M \to M_ k$ is a homomorphism such that

commutes. Let $\alpha : M \to M$ be the composition $f_ k \circ z_ k : M \to M_ k \to M$. Then $\alpha $ factors through a finitely presented module by construction and $\alpha \circ f_ j = f_ j$. Since the image of $f$ is contained in the image of $f_ j$ this also implies that $\alpha \circ f = f$. $\square$

We will see later (see Lemma 10.148.13) that Lemma 10.91.2 means that a countably generated Mittag-Leffler module over a henselian local ring is a direct sum of finitely presented modules.

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