Let $X$, $S$ be integral quasi-projective schemes (over $\Bbb C$). Let $\mathcal F$ be a coherent sheaf on $X\times S$, flat on $S$. Suppose that $x\in X$, $s\in S$ are closed points, and ${\mathcal F}(s)={\mathcal F}_{\rvert X\times\{s\}}$. Suppose that ${\mathcal F}(s)_x$ is a free ${\mathcal O}_{X,x}$-module. Does this imply that ${\mathcal F}_{x,s}$ is a free ${\mathcal O}_{X\times S,(x,s)}$-module ?
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- $\begingroup$ This seems to be true for me because of Theorem 37.16.1 in stacks.math.columbia.edu/tag/039A Set $X$ and $Y$ in the theorem both equal to $X \times_k S$, set $S$ equal to $S$ and set $\mathcal{F}$ equal to $\mathcal{F}$. Set $x$ to $(x,s)$ and $y = (x,s)$ and $s = s$. $\endgroup$Jürgen Böhm– Jürgen Böhm2022-11-05 20:10:07 +00:00Commented Nov 5, 2022 at 20:10
- 1$\begingroup$ Since $(\mathcal{F}|_{X\times\{s\}})_x=\mathcal{F}_{x,s}\otimes_{\mathcal{O}_{X\times S,(x,s)}}\mathcal{O}_{X,x}$ and $\mathcal{O}_{X\times S,(x,s)}$ is flat over $\mathcal{O}_{X,x}$, this seems like a commutative algebra problem. $\endgroup$user493118– user4931182022-11-05 20:13:46 +00:00Commented Nov 5, 2022 at 20:13
- $\begingroup$ @JürgenBöhm thank you, this works $\endgroup$Hephaistos– Hephaistos2022-11-06 08:25:01 +00:00Commented Nov 6, 2022 at 8:25
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