## 17.14 Locally free sheaves

Let $(X, \mathcal{O}_ X)$ be a ringed space. Our conventions allow (some of) the stalks $\mathcal{O}_{X, x}$ to be the zero ring. This means we have to be a little careful when defining the rank of a locally free sheaf.

Definition 17.14.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules.

1. We say $\mathcal{F}$ is locally free if for every point $x \in X$ there exist a set $I$ and an open neighbourhood $x \in U \subset X$ such that $\mathcal{F}|_ U$ is isomorphic to $\bigoplus _{i \in I} \mathcal{O}_ X|_ U$ as an $\mathcal{O}_ X|_ U$-module.

2. We say $\mathcal{F}$ is finite locally free if we may choose the index sets $I$ to be finite.

3. We say $\mathcal{F}$ is finite locally free of rank $r$ if we may choose the index sets $I$ to have cardinality $r$.

A finite direct sum of (finite) locally free sheaves is (finite) locally free. However, it may not be the case that an infinite direct sum of locally free sheaves is locally free.

Lemma 17.14.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is locally free then it is quasi-coherent.

Proof. Omitted. $\square$

Lemma 17.14.3. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. If $\mathcal{G}$ is a locally free $\mathcal{O}_ Y$-module, then $f^*\mathcal{G}$ is a locally free $\mathcal{O}_ X$-module.

Proof. Omitted. $\square$

Lemma 17.14.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Suppose that the support of $\mathcal{O}_ X$ is $X$, i.e., all stalks of $\mathcal{O}_ X$ are nonzero rings. Let $\mathcal{F}$ be a locally free sheaf of $\mathcal{O}_ X$-modules. There exists a locally constant function

$\text{rank}_\mathcal {F} : X \longrightarrow \{ 0, 1, 2, \ldots \} \cup \{ \infty \}$

such that for any point $x \in X$ the cardinality of any set $I$ such that $\mathcal{F}$ is isomorphic to $\bigoplus _{i\in I} \mathcal{O}_ X$ in a neighbourhood of $x$ is $\text{rank}_\mathcal {F}(x)$.

Proof. Under the assumption of the lemma the cardinality of $I$ can be read off from the rank of the free module $\mathcal{F}_ x$ over the nonzero ring $\mathcal{O}_{X, x}$, and it is constant in a neighbourhood of $x$. $\square$

Lemma 17.14.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $r \geq 0$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a map of finite locally free $\mathcal{O}_ X$-modules of rank $r$. Then $\varphi$ is an isomorphism if and only if $\varphi$ is surjective.

Proof. Assume $\varphi$ is surjective. Pick $x \in X$. There exists an open neighbourhood $U$ of $x$ such that both $\mathcal{F}|_ U$ and $\mathcal{G}|_ U$ are isomorphic to $\mathcal{O}_ U^{\oplus r}$. Pick lifts of the free generators of $\mathcal{G}|_ U$ to obtain a map $\psi : \mathcal{G}|_ U \to \mathcal{F}|_ U$ such that $\varphi |_ U \circ \psi = \text{id}$. Hence we conclude that the map $\Gamma (U, \mathcal{F}) \to \Gamma (U, \mathcal{G})$ induced by $\varphi$ is surjective. Since both $\Gamma (U, \mathcal{F})$ and $\Gamma (U, \mathcal{G})$ are isomorphic to $\Gamma (U, \mathcal{O}_ U)^{\oplus r}$ as an $\Gamma (U, \mathcal{O}_ U)$-module we may apply Algebra, Lemma 10.16.4 to see that $\Gamma (U, \mathcal{F}) \to \Gamma (U, \mathcal{G})$ is injective. This finishes the proof. $\square$

Lemma 17.14.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. If all stalks $\mathcal{O}_{X, x}$ are local rings, then any direct summand of a finite locally free $\mathcal{O}_ X$-module is finite locally free.

Proof. Assume $\mathcal{F}$ is a direct summand of the finite locally free $\mathcal{O}_ X$-module $\mathcal{H}$. Let $x \in X$ be a point. Then $\mathcal{H}_ x$ is a finite free $\mathcal{O}_{X, x}$-module. Because $\mathcal{O}_{X, x}$ is local, we see that $\mathcal{F}_ x \cong \mathcal{O}_{X, x}^{\oplus r}$ for some $r$, see Algebra, Lemma 10.78.2. By Lemma 17.11.6 we see that $\mathcal{F}$ is free of rank $r$ in an open neighbourhood of $x$. (Note that $\mathcal{F}$ is of finite presentation as a summand of $\mathcal{H}$.) $\square$

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