Hartshorne Problem II.5.7.
Let X be a noetherian scheme, and let $\mathcal{F}$ be a coherent sheaf.
(a) If the stalk $\mathcal{F}_x$ is a free $\OO_{X,x}$-module for some point $x\in X$, then there is a neighborhood $U$ of $x$ such that $\mathcal{F}|_U$is free.
(b) $\mathcal{F}$ is locally free if and only if its stalks $\mathcal{F}_x$ are free $\OO_{X,x}$-modules for all $x\in X$.
(c) $\mathcal{F}$ is invertible (i.e.,locally free of rank 1) if and only if there is a coherent sheaf $\mathcal{G}$ such that $\mathcal{F}\otimes\mathcal{G}\cong\OO_X$
(a) This question is local so we may assume $X=\spec A$ is affine, $\mathcal{F}\cong \tilde M$ for some finitely generated A-Module $M$. We know that for some prime $\mathfrak{p}\subset A$, $M_\mathfrak{p}\cong A_\mathfrak{p}^n$ for some $n\ge0$. If $n=0$ then $M_\mathfrak{p}=0$ and as $\op{Supp}(\mathcal{F})$ is open, there is a whole neighborhood over which $\mathcal{F}$ restricts to be $0$. Thus assume $n\ge 1$. Let $t_1,…,t_n\in M_\mathfrak{p}$ be the free generators. For each $1\le i\le n$ let $f_i\in A-\mathfrak{p}$ , $s_i\in M[\frac{1}{f_i}]$ such that the stalk of $s_i$ at $\mathfrak{p}$ is $t_i$. Set $f:=f_1\cdot …\cdot f_n$ and consider each $s_i$ as its restriction to $\{f\ne 0\}$. So, we obtain a homomorphism
$$A[\frac 1f]^{n}\xrightarrow{\varphi} {M[\frac{1}{f}]}$$
by $1_i\mapsto s_i$. It is an isomorphism at $\mathfrak{p}$ so $\op{ker}(\varphi)_{\mathfrak{p}}=\op{coker}(\varphi)_\mathfrak{p}=0$. Since $M$ is finitely generated, so is $M[\frac{1}{f}]$ and so is $\op{coker}(\varphi)$. Since $A[\frac{1}{f}]$ is noetherian so is $\op{ker}(\varphi)$ (a sub-module of a noetherian module is finitely generated). There are elements in $A-\mathfrak{p}$ that annihilate all of their generators; taking their product $g$, gives yields an open neighborhood $\{fg\ne 0\}$ of $\mathfrak{p}$ over which $\varphi$ is an isomorphism, and clearly yields an isomorphism of $O_{\{fg\ne0\}}$-modules (indeed it is one at every stalk),
$$\OO_{\{fg\ne 0\}}^{\oplus n} \cong \widetilde{M[\frac{1}{fg}]}\cong\mathcal{F}|_{\{fg\ne 0\}}.$$
(b) The forward direction is trivial from the definition and the converse follows immediately from part (a). Note the importance of the coherence of $\mathcal{F}$.
(c) If $\mathcal{F}$ is invertible, then $\mathcal{F}^\vee$ will do. Indeed $\mathcal{F}\otimes\mathcal{F}^\vee\cong\OO_X$ is easily checked on stalks, and by part (b) the same goes for showing that $\mathcal{F}^\vee$ also invertible. For the converse, assume that there is a coherent sheaf $\mathcal{G}$ such that $\mathcal{F}\otimes\mathcal{G}\cong\OO_X$. It suffices to show that $\mathcal{F}_x$ is a free $\OO_{X,x}$-module of rank 1 for every $x\in X$ (as we saw above, a coherent and locally free $\OO_X$-module has locally constant rank). On all stalks we have $\mathcal{F}_x\otimes\mathcal{G}_x\cong\OO_{X,x}$ and tensoring with $k(x)$ gives
$$ (\mathcal{F}_x\otimes_{\OO_{X,x}}k(x))\otimes_{k(x)}(\mathcal{G}_x\otimes_{\OO_{X,x}}k(x))\cong k(x).$$
Everything involved is a vector space, so taking dimensions of both sides yields that $\mathcal{F}_x\otimes_{\OO_{X,x}}k(x)$ and $\mathcal{F}_x\otimes_{\OO_{X,x}}k(x)$ are 1-dimensional $k(x)$-vector spaces. By Nakayama’s lemma, $\mathcal{F}_x$, $\mathcal{G}_x$ are generated by a single element. It remains to check that this element $m$ is not $\OO_{X,x}$-torsion. Let $m’$ be a section on an open affine neighborhood $U$ of $x$ whose stalk at $x$ is $m$. Since we always have $x\in \op{Supp}(\OO_X)$, both $\mathcal{F}_x$ and $\mathcal{G}_x$ are always nonzero, so $x\in\op{Supp}(\mathcal{F})=\op{Supp}(m’)$. Writing $\mathcal{F}|_U\cong \tilde M$ we know $\op{Ann}_A(m’)\subset x$ and it follows that $m$ is not $\OO_{X,x}$-torsion. The same is true for $\mathcal{G}$ too, which shows that both $\mathcal{F}$ and $\mathcal{G}$ are invertible.
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