
Lemma 10.148.13. Let $R$ be a henselian local ring. Any countably generated Mittag-Leffler module over $R$ is a direct sum of finitely presented $R$-modules.

Proof. Let $M$ be a countably generated and Mittag-Leffler $R$-module. We claim that for any element $x \in M$ there exists a direct sum decomposition $M = N \oplus K$ with $x \in N$, the module $N$ finitely presented, and $K$ Mittag-Leffler.

Suppose the claim is true. Choose generators $x_1, x_2, x_3, \ldots$ of $M$. By the claim we can inductively find direct sum decompositions

$M = N_1 \oplus N_2 \oplus \ldots \oplus N_ n \oplus K_ n$

with $N_ i$ finitely presented, $x_1, \ldots , x_ n \in N_1 \oplus \ldots \oplus N_ n$, and $K_ n$ Mittag-Leffler. Repeating ad infinitum we see that $M = \bigoplus N_ i$.

We still have to prove the claim. Let $x \in M$. By Lemma 10.91.2 there exists an endomorphism $\alpha : M \to M$ such that $\alpha$ factors through a finitely presented module, and $\alpha (x) = x$. Say $\alpha$ factors as

$\xymatrix{ M \ar[r]^\pi & P \ar[r]^ i & M }$

Set $a = \pi \circ \alpha \circ i : P \to P$, so $i \circ a \circ \pi = \alpha ^3$. By Lemma 10.15.2 there exists a monic polynomial $P \in R[T]$ such that $P(a) = 0$. Note that this implies formally that $\alpha ^2 P(\alpha ) = 0$. Hence we may think of $M$ as a module over $R[T]/(T^2P)$. Assume that $x \not= 0$. Then $\alpha (x) = x$ implies that $0 = \alpha ^2P(\alpha )x = P(1)x$ hence $P(1) = 0$ in $R/I$ where $I = \{ r \in R \mid rx = 0\}$ is the annihilator of $x$. As $x \not= 0$ we see $I \subset \mathfrak m_ R$, hence $1$ is a root of $\overline{P} = P \bmod \mathfrak m_ R \in R/\mathfrak m_ R[T]$. As $R$ is henselian we can find a factorization

$T^2P = (T^2 Q_1) Q_2$

for some $Q_1, Q_2 \in R[T]$ with $Q_2 = (T - 1)^ e \bmod \mathfrak m_ R R[T]$ and $Q_1(1) \not= 0 \bmod \mathfrak m_ R$, see Lemma 10.148.3. Let $N = \mathop{\mathrm{Im}}(\alpha ^2Q_1(\alpha ) : M \to M)$ and $K = \mathop{\mathrm{Im}}(Q_2(\alpha ) : M \to M)$. As $T^2Q_1$ and $Q_2$ generate the unit ideal of $R[T]$ we get a direct sum decomposition $M = N \oplus K$. Moreover, $Q_2$ acts as zero on $N$ and $T^2Q_1$ acts as zero on $K$. Note that $N$ is a quotient of $P$ hence is finitely generated. Also $x \in N$ because $\alpha ^2Q_1(\alpha )x = Q_1(1)x$ and $Q_1(1)$ is a unit in $R$. By Lemma 10.88.10 the modules $N$ and $K$ are Mittag-Leffler. Finally, the finitely generated module $N$ is finitely presented as a finitely generated Mittag-Leffler module is finitely presented, see Example 10.90.1 part (1). $\square$

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