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 10.92.2. Lemma 10.92.1 does not hold without the countable generation assumption. For example, the $\mathbf Z$-module $M = \mathbf{Z}[[x]]$ is flat and Mittag-Leffler but not projective. It is Mittag-Leffler by Lemma 10.90.4. Subgroups of free abelian groups are free, hence a projective $\mathbf Z$-module is in fact free and so are its submodules. Thus to show $M$ is not projective it suffices to produce a non-free submodule. Fix a prime $p$ and consider the submodule $N$ consisting of power series $f(x) = \sum a_ i x^ i$ such that for every integer $m \geq 1$, $p^ m$ divides $a_ i$ for all but finitely many $i$. Then $\sum a_ i p^ i x^ i$ is in $N$ for all $a_ i \in \mathbf{Z}$, so $N$ is uncountable. Thus if $N$ were free it would have uncountable rank and the dimension of $N/pN$ over $\mathbf{Z}/p$ would be uncountable. This is not true as the elements $x^ i \in N/pN$ for $i \geq 0$ span $N/pN$.


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