Lemma 48.29.1. Let $X$ be a locally ringed space. Let

$\mathcal{E}_1 \xrightarrow {\alpha } \mathcal{E}_0 \to \mathcal{F} \to 0$

be a short exact sequence of $\mathcal{O}_ X$-modules. Assume $\mathcal{E}_1$ and $\mathcal{E}_0$ are locally free of ranks $r_1, r_0$. Then there is a canonical map

$\wedge ^{r_0 - r_1}\mathcal{F} \longrightarrow \wedge ^{r_1}(\mathcal{E}_1^\vee ) \otimes \wedge ^{r_0}\mathcal{E}_0$

which is an isomorphism on the stalk at $x \in X$ if and only if $\mathcal{F}$ is locally free of rank $r_0 - r_1$ in an open neighbourhood of $x$.

Proof. If $r_1 > r_0$ then $\wedge ^{r_0 - r_1}\mathcal{F} = 0$ by convention and the unique map cannot be an isomorphism. Thus we may assume $r = r_0 - r_1 \geq 0$. Define the map by the formula

$s_1 \wedge \ldots \wedge s_ r \mapsto t_1^\vee \wedge \ldots \wedge t_{r_1}^\vee \otimes \alpha (t_1) \wedge \ldots \wedge \alpha (t_{r_1}) \wedge \tilde s_1 \wedge \ldots \wedge \tilde s_ r$

where $t_1, \ldots , t_{r_1}$ is a local basis for $\mathcal{E}_1$, correspondingly $t_1^\vee , \ldots , t_{r_1}^\vee$ is the dual basis for $\mathcal{E}_1^\vee$, and $s'_ i$ is a local lift of $s_ i$ to a section of $\mathcal{E}_0$. We omit the proof that this is well defined.

If $\mathcal{F}$ is locally free of rank $r$, then it is straightforward to verify that the map is an isomorphism. Conversely, assume the map is an isomorphism on stalks at $x$. Then $\wedge ^ r\mathcal{F}_ x$ is invertible. This implies that $\mathcal{F}_ x$ is generated by at most $r$ elements. This can only happen if $\alpha$ has rank $r$ modulo $\mathfrak m_ x$, i.e., $\alpha$ has maximal rank modulo $\mathfrak m_ x$. This implies that $\alpha$ has maximal rank in a neighbourhood of $x$ and hence $\mathcal{F}$ is locally free of rank $r$ in a neighbourhood as desired. $\square$

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