The Stacks project

Proof. We will prove this by induction on $n$.

Base case: $n = 0$. Then $P_ e = M_ e/\bigcup _{e' < e} M_{e'}$ is projective. Thus we may choose a section $P_ e \to M_ e$ of the projection $M_ e \to P_ e$. We claim that the induced map $\psi : \bigoplus _{e \in E} P_ e \to M$ is an isomorphism. Namely, if $x = \sum x_ e \in \bigoplus P_ e$ is nonzero, then we let $e_{max}$ be maximal such that $x_{e_{max}}$ is nonzero and we conclude that $y = \psi (x) = \psi (\sum x_ e)$ is nonzero because $y \in M_{e_{max}}$ has nonzero image $x_{e_{max}}$ in $P_{e_{max}}$. On the other hand, let $y \in M$. Then $y \in M_ e$ for some $e$. We show that $y \in \mathop{\mathrm{Im}}(\psi )$ by transfinite induction on $e$. Let $x_ e \in P_ e$ be the image of $y$. Then $y - \psi (x_ e) \in \bigcup _{e' < e} M_{e'}$. By induction hypothesis we conclude that $y - \psi (x_ e) \in \mathop{\mathrm{Im}}(\psi )$ hence $y \in \mathop{\mathrm{Im}}(\psi )$. Thus the claim is true and $\psi$ is an isomorphism. We conclude that $M$ is projective as a direct sum of projectives, see Lemma 10.77.4.

If $n > 0$, then for $e \in E$ we denote $F_ e$ the free $R$-module on the set of elements of $M_ e$. Then we have a system of short exact sequences

over the well-ordered set $E$. Note that the transition maps $F_{e'} \to F_ e$ and $K_{e'} \to K_ e$ are injective too. Set $F = \bigcup F_ e$ and $K = \bigcup K_ e$. Then

is a short exact sequence of $R$-modules too and $F_ e/\bigcup _{e' < e} F_{e'}$ is the free $R$-module on the set of elements in $M_ e$ which are not contained in $\bigcup _{e' < e} M_{e'}$. Hence by Lemma 10.109.9 we see that the projective dimension of $K_ e/\bigcup _{e' < e} K_{e'}$ is at most $n - 1$. By induction we conclude that $K$ has projective dimension at most $n - 1$. Whence $M$ has projective dimension at most $n$ and we win. $\square$