# Extending a property of coherent sheaf

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?

• Hint: consider the rank function of $\mathcal{G}$ as a sheaf on $X$. It is upper semi-continuous and integer valued, so... – KReiser Sep 24 '18 at 4:50
• @KReiser: ...this function is locally constant? – Tim Grosskreutz Sep 24 '18 at 20:22
• 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? – KReiser Sep 24 '18 at 20:48
• @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. – Tim Grosskreutz Sep 24 '18 at 23:12
• 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. – KReiser Sep 24 '18 at 23:37