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 10.76.2. Let $R$ be a ring. Let $P$ be an $R$-module. The following are equivalent

  1. $P$ is projective,

  2. $P$ is a direct summand of a free $R$-module, and

  3. $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ for every $R$-module $M$.

Proof. Assume $P$ is projective. Choose a surjection $\pi : F \to P$ where $F$ is a free $R$-module. As $P$ is projective there exists a $i \in \mathop{\mathrm{Hom}}\nolimits _ R(P, F)$ such that $i \circ \pi = \text{id}_ P$. In other words $F \cong \mathop{\mathrm{Ker}}(\pi ) \oplus i(P)$ and we see that $P$ is a direct summand of $F$.

Conversely, assume that $P \oplus Q = F$ is a free $R$-module. Note that the free module $F = \bigoplus _{i \in I} R$ is projective as $\mathop{\mathrm{Hom}}\nolimits _ R(F, M) = \prod _{i \in I} M$ and the functor $M \mapsto \prod _{i \in I} M$ is exact. Then $\mathop{\mathrm{Hom}}\nolimits _ R(F, -) = \mathop{\mathrm{Hom}}\nolimits _ R(P, -) \times \mathop{\mathrm{Hom}}\nolimits _ R(Q, -)$ as functors, hence both $P$ and $Q$ are projective.

Assume $P \oplus Q = F$ is a free $R$-module. Then we have a free resolution $F_\bullet $ of the form

\[ \ldots F \xrightarrow {a} F \xrightarrow {b} F \to P \to 0 \]

where the maps $a, b$ alternate and are equal to the projector onto $P$ and $Q$. Hence the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , M)$ is split exact in degrees $\geq 1$, whence we see the vanishing in (3).

Assume $\mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M) = 0$ for every $R$-module $M$. Pick a free resolution $F_\bullet \to P$. Set $M = \mathop{\mathrm{Im}}(F_1 \to F_0) = \mathop{\mathrm{Ker}}(F_0 \to P)$. Consider the element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ R(P, M)$ given by the class of the quotient map $\pi : F_1 \to M$. Since $\xi $ is zero there exists a map $s : F_0 \to M$ such that $\pi = s \circ (F_1 \to F_0)$. Clearly, this means that

\[ F_0 = \mathop{\mathrm{Ker}}(s) \oplus \mathop{\mathrm{Ker}}(F_0 \to P) = P \oplus \mathop{\mathrm{Ker}}(F_0 \to P) \]

and we win. $\square$


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