Remark 10.88.13. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module which is Mittag-Leffler as an $R$-module. Then it is in general not the case that $M$ is Mittag-Leffler as an $S$-module. For example suppose that $S$ is the ring of dual numbers over $R$, i.e., $S = R \oplus R\epsilon$ with $\epsilon ^2 = 0$. Then an $S$-module consists of an $R$-module $M$ endowed with a square zero $R$-linear endomorphism $\epsilon : M \to M$. Now suppose that $M_0$ is an $R$-module which is not Mittag-Leffler. Choose a presentation $F_1 \xrightarrow {u} F_0 \to M_0 \to 0$ with $F_1$ and $F_0$ free $R$-modules. Set $M = F_1 \oplus F_0$ with

$\epsilon = \left( \begin{matrix} 0 & 0 \\ u & 0 \end{matrix} \right) : M \longrightarrow M.$

Then $M/\epsilon M \cong F_1 \oplus M_0$ is not Mittag-Leffler over $R = S/\epsilon S$, hence not Mittag-Leffler over $S$ (see Lemma 10.88.12). On the other hand, $M/\epsilon M = M \otimes _ S S/\epsilon S$ which would be Mittag-Leffler over $S$ if $M$ was, see Lemma 10.88.9.

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