Example 10.91.5. Non Mittag-Leffler modules.
By Example 10.89.1 and Proposition 10.89.5, \mathbf{Q} is not a Mittag-Leffler \mathbf{Z}-module.
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.
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.
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).
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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