Let $\pi: X \to C$ be a fibration (proper, flat with generically smooth irreducible equidimensional fibers) over smooth irreducible $k$-scheme $C$, $k$ field, eg $C$ curve. Let $L$ be a line bundle on $X$ which is assumed to be trivial over generic fiber $X_{\eta}$.

What are the mildest additional assumpions on $X$ and $C$ should be required to guarantee the existence an open $U \subset C$ such that $L$ trivializes over it's preimage $\pi^{-1}(U)$?

A remark on the smoothness / geom. regular assumptions of the base $C$ respectively the fibers: is this assumption really neccessary or does it suffice to require for $C$ to be regular without worsening the situation?

  • $\begingroup$ Is everything happening over a field here, or are you using the smooth in "smooth irreducible base scheme" to mean a regular scheme? I'm asking because you're asking for mildest conditions and it probably matters. $\endgroup$
    – KReiser
    Dec 5, 2022 at 18:05
  • $\begingroup$ @KReiser: yes, it should be reasonable to assume that $C$ is a $k$-scheme, $k$ field $\endgroup$
    – user267839
    Dec 5, 2022 at 18:09
  • $\begingroup$ so $C$ smooth or equivalently geometrically regular in order to avoid troubles with non perfect fields... $\endgroup$
    – user267839
    Dec 5, 2022 at 18:17
  • $\begingroup$ Or can the assumption be weakended to $C$ regular over $k$? The idea I had in mind was to extend a nowhere in $X_{\eta}$ vanishing section $s \in H^0(X_{\eta}, L_{\eta})= H^0(X_{\eta})$ to a section $\hat{s}$ of $L$ in an open $U \subset X$ containing the generic fiber. Then it would vanish in a divisor $V_s$ of $X$ disjoint from $X_{\eta}$. restricting $\hat{s}$ to intersection of $U$ with the complement of $V_s$ would give an open subscheme containing the generic fiber where $\hat{s}$ trivializes $L$. $\endgroup$
    – user267839
    Dec 5, 2022 at 18:40
  • $\begingroup$ We still have to show that this open subscheme contains a preimage of an open subscheme $V \subset C$, but this looks doable. So if I haven't made any mistakes anywhere, the main hurdle is to lift $s$ to an $U$. does it make sense, or could it be approached more sophisticatedly? $\endgroup$
    – user267839
    Dec 5, 2022 at 18:47

2 Answers 2


By restricting to an open set in $ C $, we can and will assume that all fibers are smooth and irreducible. The locus of points $ \{ p \in C; L|_{X_p} \cong \mathcal{O}_{X_p} \} $ is closed by semicontinuity - it is given by the intersection of two closed sets $$ \{ p \in C; \dim H^0( X_p, L|_{X_p} ) > 0 \} \cap \{ p \in C; \dim H^0( X_p, L^{\vee}|_{X_p} ) > 0 \} $$ and by assumption the locus contains the generic point, so is all of $ C $.

Because $ L $ is trivial on every fiber, $ \pi_* L $ is a line bundle by cohomology and base change. Let $ U \subset C $ be an open set where $ \pi_* L $ is trivial. Then $ L $ is trivial on $ \pi^{-1} (U) $. To see this, it’s enough to see that $ \pi^* \pi_* L $ is trivial there (being isomorphic to $ L $) and that’s obvious from construction.

  • $\begingroup$ why fiberwise trivialness of $ L $ implies that $ \pi_* L $ is line bundle? You say that it follows from cohomology and base change. Is it some standard result? could you expain to which result there you are refering to? $\endgroup$
    – user267839
    Dec 6, 2022 at 15:38
  • $\begingroup$ @JustusC: See for instance these notes: math.uchicago.edu/~amathew/semicontinuity.pdf . $\endgroup$
    – Aphelli
    Dec 6, 2022 at 15:44

The existing answer is good, but I wanted to give a slightly more elementary (and, I feel, more natural) argument, under very weak assumptions (here, $C$ is an irreducible Noetherian scheme and $X$ is of finite type over $C$). We can obviously assume that $C$ is affine, and thus $X$ is quasi-compact.

Edit: I think we also need to assume that $O_{C,\eta}$ (where $\eta \in C$ is the generic point) is reduced (can you see where it is needed?).

Note that Chevalley’s theorem says the following: if $A \subset X$ is a locally closed subset not meeting the generic fiber of $\pi$, then it doesn’t meet $\pi^{-1}(V)$ for some nonempty open subset $V \subset C$ (consider the image of $A$ in $C$, which is constructible but doesn’t contain the generic point).

I claim that there exists an open subset $U \subset X$ containing the generic fiber and a $s \in H^0(U,L)$ which is trivializing on the generic fiber.

Proof: let $X_{\eta}$ be the generic fiber, and $\sigma \in H^0(X_{\eta},L)$ be a trivializing section. We can find finitely many open subsets $U_i \subset X$ such that they cover $X_{\eta}$, and sections $s_i \in H^0(U_i,L)$ such that $(s_i)_{|U_i\cap X_{\eta}}=\sigma_{|U_i \cap X_{\eta}}$.

For every pair $i \neq j$, consider $D_{i,j}$ to be the locally closed (in $X$) subset of $x \in U_i \cap U_j$ such that $(s_i)_{x}\neq (s_j)_x$. The union $D$ of the $D_{i,j}$ is a locally closed subset of $X$ and does not meet the generic fiber, so by Chevalley it doesn’t meet some $\pi^{—1}(W)$ where $W \subset C$ is some nonempty open subset.

Therefore, $X_W$ contains no point in any of the $D_{i,j}$. It follows that the $(s_i)_{|X_W \cap U_i}$ glue to a section $s \in H^0(X_W \cap \cup_i{U_i},L)$.

How do we conclude? Let $s,U$ be as above, and let $D(s)=\{x \in U,\, s_x \notin m_{X,x}L_x\}$. It’s easy to see that $D(s)$ is open in $X$, so its complement is closed and doesn’t meet $X_{\eta}$. By Chevalley, there is a nonempty proper open subset $V \subset C$ such that $\pi^{-1}(V)$ doesn’t meet $X \backslash D(s)$; thus $D(s) \supset \pi^{-1}(V)$ and thus $L$ is free over $\pi^{-1}(V)$ with basis $s$.

Other Edit: the claim has a simpler proof when $O_{C,\eta}$ is reduced and $\pi$ is qcqs. Then $L’=\pi_{\ast}(L)$ is a quasi-coherent $O_C$-module; moreover, by flat base change the natural map $L’_{\eta} \rightarrow H^0(X_{\eta},L)$ is an isomorphism). It follows that $s$ comes from a section of $L’_{\eta}$, thus from a section of $L’(U)=H^0(\pi^{—1}(U),L)$ for some nonempty open subset $U \subset L$.

  • $\begingroup$ Yes, this is the more geometric, and better answer. $\endgroup$ Dec 6, 2022 at 15:29
  • $\begingroup$ It looks like I got carried away and stated a result with a bit too much generality. I think this kind of geometric argument can’t escape the constraint that $O_{C,\eta}$ be reduced – otherwise, there’s a significant difference between taking the generic fiber (a “bad” operation, mainly because it’s not flat) and tensoring with $O_{C,\eta}$ (which is good, because it’s the kind of operation that actually extends to an open neighborhood, and it is flat). $\endgroup$
    – Aphelli
    Dec 7, 2022 at 13:29

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