Lemma 17.18.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a flat $\mathcal{O}_ X$-module of finite presentation. Then $\mathcal{F}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module.

Proof. After replacing $X$ by the members of an open covering, we may assume there exists a presentation

$\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{F} \to 0$

Let $x \in X$. By Lemma 17.17.10 we can, after shrinking $X$ to an open neighbourhood of $x$, assume there exists a factorization

$\mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1} \to \mathcal{F}$

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1}$ annihilates the first summand of $\mathcal{O}_ X^{\oplus r}$. Repeating this argument $r - 1$ more times we obtain a factorization

$\mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F}$

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r}$ is zero. This means that the surjection $\mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F}$ has a section and we win. $\square$

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