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*}

Lemma 10.87.10. Let $R$ be a ring and $M$ an $R$-module. Then $M$ is Mittag-Leffler if and only if for every finite free $R$-module $F$ and module map $f: F \to M$, there exists a finitely presented $R$-module $Q$ and a module map $g : F \to Q$ such that $g$ and $f$ dominate each other, i.e., $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ N) = \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ N)$ for every $R$-module $N$.

Proof. Since the condition is clear weaker than condition (1) of Proposition 10.87.6 we see that a Mittag-Leffler module satisfies the condition. Conversely, suppose that $M$ satisfies the condition and that $f : P \to M$ is an $R$-module map from a finitely presented $R$-module $P$ into $M$. Choose a surjection $F \to P$ where $F$ is a finite free $R$-module. By assumption we can find a map $F \to Q$ where $Q$ is a finitely presented $R$-module such that $F \to Q$ and $F \to M$ dominate each other. In particular, the kernel of $F \to Q$ contains the kernel of $F \to P$, hence we obtain an $R$-module map $g : P \to Q$ such that $F \to Q$ is equal to the composition $F \to P \to Q$. Let $N$ be any $R$-module and consider the commutative diagram

\[ \xymatrix{ F \otimes _ R N \ar[d] \ar[r] & Q \otimes _ R N \\ P \otimes _ R N \ar[ru] \ar[r] & M \otimes _ R N } \]

By assumption the kernels of $F \otimes _ R N \to Q \otimes _ R N$ and $F \otimes _ R N \to M \otimes _ R N$ are equal. Hence, as $F \otimes _ R N \to P \otimes _ R N$ is surjective, also the kernels of $P \otimes _ R N \to Q \otimes _ R N$ and $P \otimes _ R N \to M \otimes _ R N$ are equal. $\square$

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 05CP. Beware of the difference between the letter 'O' and the digit '0'.