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\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 10.108.3. Let $R$ be a ring. Suppose that $M$ is an $R$-module of projective dimension $d$. Suppose that $F_ e \to F_{e-1} \to \ldots \to F_0 \to M \to 0$ is exact with $F_ i$ projective and $e \geq d - 1$. Then the kernel of $F_ e \to F_{e-1}$ is projective (or the kernel of $F_0 \to M$ is projective in case $e = 0$).

Proof. We prove this by induction on $d$. If $d = 0$, then $M$ is projective. In this case there is a splitting $F_0 = \mathop{\mathrm{Ker}}(F_0 \to M) \oplus M$, and hence $\mathop{\mathrm{Ker}}(F_0 \to M)$ is projective. This finishes the proof if $e = 0$, and if $e > 0$, then replacing $M$ by $\mathop{\mathrm{Ker}}(F_0 \to M)$ we decrease $e$.

Next assume $d > 0$. Let $0 \to P_ d \to P_{d-1} \to \ldots \to P_0 \to M \to 0$ be a minimal length finite resolution with $P_ i$ projective. According to Schanuel's Lemma 10.108.1 we have $P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \cong F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$. This proves the case $d = 1$, $e = 0$, because then the right hand side is $F_0 \oplus P_1$ which is projective. Hence now we may assume $e > 0$. The module $F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$ has the finite projective resolution

\[ 0 \to P_ d \to P_{d-1} \to \ldots \to P_2 \to P_1 \oplus F_0 \to \mathop{\mathrm{Ker}}(P_0 \to M) \oplus F_0 \to 0 \]

of length $d - 1$. By induction applied to the exact sequence

\[ F_ e \to F_{e-1} \to \ldots \to F_2 \to P_0 \oplus F_1 \to P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \to 0 \]

of length $e - 1$ we conclude $\mathop{\mathrm{Ker}}(F_ e \to F_{e - 1})$ is projective (if $e \geq 2$) or that $\mathop{\mathrm{Ker}}(F_1 \oplus P_0 \to F_0 \oplus P_0)$ is projective. This implies the lemma. $\square$


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