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Let $C$ be a regular projective integer curve over a field $k$ and $X\rightarrow C$ a projective $k$-Variety with generic fibre $X_\eta$. Why is there for every line bundle (invertible sheaf) $L$ on $X_\eta$ a nonempty open subset $U\subseteq C$ and a line bundle $\mathcal{L}$ on $X_U:=X\times_CU$ with $\mathcal{L}|_{X_\eta}=L$?

And if there's another open subset $U'\subseteq C$ and another line bundle $\mathcal{L'}$ satisfying $\mathcal{L}'|_{X_\eta}=L$, is there an open subset $W\subseteq U\cap U'$ with $\mathcal{L}|_W=\mathcal{L}'|_W$?

Edit: Previously I had the setting $U\subseteq X$, but now there's the right question.

Altinior
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2 Answers2

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Since nobody has answered this yet, let me give a sketch of an argument. (I am sure it can be made more elegant, but I think the essential idea is ok.)

First of all we can assume $C$ is affine (of any dimension), say $C=\operatorname{Spec} A$ for some f.g $k$-algebra $A$. Then $X = \operatorname{Proj} (B)$ for some f.g. graded $A$-algebra $B$.

Instead of talking about line bundles, let me talk about Cartier divisors instead. I'll show how to extend a Cartier divisor on $X_\eta$, and the same idea will show how to extend linear equivalences, so we get the result for line bundles.

Note that $X_\eta = \operatorname{Proj} (B \otimes_A K)$, where $K$ is the quotient field of $A$.

A Cartier divisor on $X_\eta$ is specified by giving data $(U_i,f_i)$ where the $U_i$ form an open cover of $X_\eta$, each $f_i$ is a regular function on $U_i$, and we have the compatibility condition that $f_i/f_j$ is invertible on $U_i \cap U_j$. Notice that since $X_\eta$ is projective (hence quasi-compact) there are finitely many open sets in the cover. This is important!

The key point now is to show that $U_i = X_\eta \cap V_i$ for an open set $V_i$ in $X \times_C U$ for $U \subset C$ some open set, and that $f_i$ extends to a regular function $g_i$ on $V_i$.

To do this, notice that $U_i$ is defined by the nonvanishing of finitely elements of $B \otimes_A K$. But these elements all belong to $B \otimes_A A_f$ for a single localisation $A_f$ of $A$, so there is an open set $V_i \subset X \otimes_C \operatorname{Spec} A_f$ such that $V_i \cap X_\eta = U_i$. Similarly, the regular function $f_i$, which is an element of some localisation of $B \otimes_A K$, in fact belongs to a localisation of $B \otimes_A A_g$ for some $g \in A$. So replacing $f$ and $g$ by $h=fg$, we get a single localisation $A_h$ of $A$ such that

  1. $U_i = V_i \cap X_\eta$ for an open subset $V_i \subset X \times_C \operatorname{Spec} A_h$

  2. $f_i$ belongs to the image of $\mathcal{O}(V_i) \rightarrow \mathcal{O}(U_i)$.

Doing this for each of the finitely many pairs $(U_i,F_i)$ defining our Cartier divisor on $X_\eta$, and intersecting the corresponding open subsets of $C$, we extend the divisor to $X \times_C U$ for an open subset $U$. (Strictly speaking there is a bit more to check, namely that the compatibility conditions still hold on the bigger set, but the same arguments will apply.)

Finally, your second question is just asking if the trivial bundle on $X_\eta$ extends to the trivial bundle over some open subset in the base. That's certainly true.

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To add to the former answer: once one has the compatible transition functions on open sets $V_i$ covering the generic fibre, $f'_i\in {\mathcal O}(V_i)$, so that $f'_i/f'_j$ is invertible on $V_i\cap V_j$, one takes $F:= X\setminus \bigcup V_i,$ which is closed and is disjoint with the generic fibre, and thus if $p:X\to C$ is the structure map, $T=p(F)$ does not contain the generic point of $C$, which means that we have extended ${\mathcal L}$ to $L\in Pic(p^{-1}(C-T))$, as desired.

The essential point is that properness (we don't need full projectivity) allows for $p$ to be a closed map, and that's how we ended the argument.

Theon Alexander
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