Lemma 56.7.8. Let $k$ be a field. Let $X$, $Y$ be finite type schemes over $k$ with $X$ separated and $Y$ reduced. If there is a $k$-linear equivalence $F : \textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ Y)$ of categories, then there is an isomorphism $f : Y \to X$ over $k$ and an invertible $\mathcal{O}_ Y$-module $\mathcal{L}$ such that $F(\mathcal{F}) = f^*\mathcal{F} \otimes \mathcal{L}$.
Weak version of the result in [Gabriel] stating that the category of quasi-coherent modules determines the isomorphism class of a scheme.
Proof using Gabriel-Rosenberg reconstruction. This lemma is a weak form of the results discussed in Lemma 56.7.2 and Remark 56.7.3. $\square$
Proof not relying on Gabriel-Rosenberg reconstruction. By Lemma 56.7.5 we obtain a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ which is flat over $X$ with support finite over $Y$ such that $F$ is given by the restriction of the functor (56.5.1.1) to $\textit{Coh}(\mathcal{O}_ X)$. If we can show that $F(\mathcal{O}_ X)$ is an invertible $\mathcal{O}_ Y$-module, then by Lemma 56.7.6 we see that $\mathcal{K} = s_*\mathcal{L}$ for some section $s : Y \to X \times Y$ of $\text{pr}_2$ and some invertible $\mathcal{O}_ Y$-module $\mathcal{L}$. This will show that $F$ has the form indicated with $f = \text{pr}_1 \circ s$. Some details omitted.
It remains to show that $F(\mathcal{O}_ X)$ is invertible. We only sketch the proof and we omit some of the details. For a closed point $x \in X$ we denote $\mathcal{O}_ x$ in $\textit{Coh}(\mathcal{O}_ X)$ the skyscraper sheaf at $x$ with value $\kappa (x)$. First we observe that the only simple objects of the category $\textit{Coh}(\mathcal{O}_ X)$ are these skyscraper sheaves $\mathcal{O}_ x$. The same is true for $Y$. Hence for every closed point $y \in Y$ there exists a closed point $x \in X$ such that $\mathcal{O}_ y \cong F(\mathcal{O}_ x)$. Moreover, looking at endomorphisms we find that $\kappa (x) \cong \kappa (y)$ as finite extensions of $k$. Then
\[ \mathop{\mathrm{Hom}}\nolimits _ Y(F(\mathcal{O}_ X), \mathcal{O}_ y) \cong \mathop{\mathrm{Hom}}\nolimits _ Y(F(\mathcal{O}_ X), F(\mathcal{O}_ x)) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, \mathcal{O}_ x) \cong \kappa (x) \cong \kappa (y) \]
This implies that the stalk of the coherent $\mathcal{O}_ Y$-module $F(\mathcal{O}_ X)$ at $y \in Y$ can be generated by $1$ generator (and no less) for each closed point $y \in Y$. It follows immediately that $F(\mathcal{O}_ X)$ is locally generated by $1$ element (and no less) and since $Y$ is reduced this indeed tells us it is an invertible module. $\square$
Comments (2)
Comment #5120 by Pieter Belmans on
Comment #5124 by Johan on