Let $f: X \to Y$ be a morphism between $k$-schemes where $Y$ is irreducible with unique generic point $\eta$. Set $F := f^{-1}(\eta)$ as the generic fiber and consider a coherent sheaf $\mathcal{G}$ on $X$.

Let assump that $\mathcal{G}$ has follwing property:

The restriction $\mathcal{G} \vert _F$ is invertible, so locally free of rank $1$.

Notice that since $F$ is generally not an open subset the "restriction" is the pullback $i_F^*\mathcal{G}$ for canonical inclusion morphism $i_F: F \to X$.

I would like to know how to prove that the property above is extendabl in follwing sense: the claim is that there exist an open $F \subset U \subset _o X$ subscheme of $X$ such that the restriction $\mathcal{G} \vert _U$ remains invertible.

My Ideas: Maybe to go pointwise in the sense by considering an arbitrary point $x \in F$ and trying to find an open neighbourhood $U_x$ of it with desired property. Then take as $U $ the union of all $U_x$. I see the advantage of this approach because we can work locally and therefore assuming that $X$ is affine. Does it work or do I need here another better approach?

  • $\begingroup$ Hint: consider the rank function of $\mathcal{G}$ as a sheaf on $X$. It is upper semi-continuous and integer valued, so... $\endgroup$ – KReiser Sep 24 '18 at 4:50
  • $\begingroup$ @KReiser: ...this function is locally constant? $\endgroup$ – Tim Grosskreutz Sep 24 '18 at 20:22
  • $\begingroup$ Well, locally constant might not be quite right, but you can pick an open neighborhood $U$ of $F$ so that the sheaf restricted to $U$ is rank 1. Can you go from here? $\endgroup$ – KReiser Sep 24 '18 at 20:48
  • $\begingroup$ @KReiser: I'm not sure. Firstly what argument exactly guarantees to me that I can pick such $U$? I don't see a different one from beeing locally constant. Concerning the next step I think that your hint refers now to exclude that $\mathcal{G} \vert _U$ has a torsion? In this case it suffice to show that the restriction of $\mathcal{G} \vert _U$ to $F$ is injective. $\endgroup$ – Tim Grosskreutz Sep 24 '18 at 23:12
  • $\begingroup$ The definition of semicontinuity with a choice of $x_0=F$ and $\epsilon = 1/2$ gives you an open $U$ so that the sheaf is rank 0 or 1 on $U$. If it's rank 0 at any point $u$ in $U$, then the same argument applied to $u$ shows that the sheaf is supported on a closed subset, which means that the sheaf is of rank 0 on the generic fiber, contradiction. Now you have a sheaf which is of the correct rank everywhere. Since freeness fails on a closed subset, this gives you a locally free sheaf of rank 1 and you are finished. $\endgroup$ – KReiser Sep 24 '18 at 23:37

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