The Stacks project

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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