The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Remark 10.87.13. 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 10.87.12). 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 10.87.9.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05CS. Beware of the difference between the letter 'O' and the digit '0'.