Let $f:X\to S$ be a smooth proper morphism of schemes with geometrically connected fibres. Assume $S$ is a smooth irreducible variety over $\mathbb{C}$. Assume that there is a sequence of closed points $(s_i)_{i=1}^\infty$ in $S$ such that the fibres $X_{s_i}$ are pairwise non-isomorphic over $\mathbb{C}$.

Is there a curve $C\subset S$ such that the restricted family $f|_{X_C}:X_C\to C$ is non-isotrivial? (That is, are there infinitely many points $c_i$ such that the fibres $X_{c_i}$ are pairwise non-isomorphic?)

I think the answer is yes, and that one can construct $C$ by taking a "general" curve. But how does one make this precise?

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    $\begingroup$ Suppose $f_{|X_C}$ is isotrivial for all curves $C\subset S$; fix a point $s$ in $S$. For every point $t\in S$ there is a curve passing through $s$ and $t$, so $X_t$ is isomorphic to $X_s$. This contradicts your hypothesis. $\endgroup$ – abx Nov 2 '17 at 10:30
  • $\begingroup$ @abx I agree with what you write. However, with the definition in the question, isotriviality doesn't a priori imply that $X_t$ and $X_s$ are isomorphic. Isotriviality just says that the isomorphism classes of the fibres $X_c$ range over a finite set. So it might still happen that $X_s$ and $X_t$ are non-isomorphic. But surely there's a way around this. $\endgroup$ – Ronnie Nov 2 '17 at 10:37
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    $\begingroup$ Where did you find this definition of isotriviality??? $\endgroup$ – abx Nov 2 '17 at 11:54

Let $k$ be an algebraically closed field that is uncountable. Let $S$ be finite type $k$-scheme. Let $f:X\to S$ be a proper, flat morphism of schemes.

Proposition. For every closed subset $T\subset S$, if there exists an infinite subset of $T(k)$ such that the corresponding fibers of $f$ are pairwise non-isomorphic, then there exists an irreducible curve $C\subset T$ such that $X_C\to C$ is non-isotrivial. In particular, for $T$ equals $S,$ if $f$ has pairwise non-isomorphic fibers over an infinite subset of $S(k)$, then there exists an irreducible curve $C\subset S$ such that $X_C\to C$ is non-isotrivial.

Proof. This is proved by Noetherian induction on the closed subset $T$ of $S.$ If $T$ is reducible, then the infinite subset of $T$ has infinite intersection with one of the finitely many irreducible components of $T.$ Thus, to prove the proposition, it suffices to assume that $T$ is irreducible.

Denote $X\times_S T$ by $X_T.$ On $T\times T,$ form the two pullback families $X_1=X_T\times T$ and $X_2 = T\times X_T.$ Edit. Thanks to Laurent Moret-Bailly who pointed out the projectivity issue. First assume that $X$ is projective. Then by the existence of the Hilbert scheme, there is a locally finite type morphism of schemes $\rho:I\to T\times T$, whose domain $I =\text{Isom}_{T\times T}(X_1,X_2)$ is a countable increasing union of quasi-compact opens $I_j$, and there exists a universal $I$-isomorphism of the pullback families, $$ \phi: I\times_{T\times T} X_1 \to I\times_{T\times T} X_2.$$ In case $f:X\to S$ is only proper, flat (and locally finitely presented), then Artin's paper Algebraization of Formal Moduli, I (reference to follow) proves that there is such $I$ that is an algebraic space. Every algebraic space has an fppf representable morphism $I'\to I$ such that $I'$ is a scheme, which means that $\phi'$ over $I'$ will not be universal, but for every isomorphism $\phi_J$ after base change by a morphism $J\to T\times T$, there will exist an fppf morphism of schemes $J'\to J$ and a morphism $J'\to I'$ such that the pullback of $\phi_J$ to $J'$ equals the pullback of $\phi'$ to $J'$. In both cases, the image of $I\to S\times S$, resp. of $I'\to S\times S$, is a countable increasing union of constructible subsets $Z_j=\rho(I_j).$

Since $k$ is uncountable, if the union contains all $k$-points of a nonempty Zariski open (or even just a nonempty analytic open when $k=\mathbb{C}$), then one of the $Z_j$ contains a dense Zariski open. In this case, the union of all opens contained in the image of $\rho$ is a dense open of the form $U\times U$ for $U\subset T$ a dense Zariski open. Thus, the infinite subset of $T$ must be in the complement $T'=T\setminus U.$ By Noetherian induction, the claim holds in this case.

Thus, assume that the image of $\rho$ contains no nonempty open, i.e., every closure $\overline{Z}_j$ is a nowhere dense closed subset of $T\times T.$ Since $k$ is uncountable, there exists a point $p\in T(k)$ such that $\{p\}\times T$ has nowhere dense intersection with every $\overline{Z}_j.$ For a very general complete intersection curve $C\subset T$ that contains $p,$ also $\{p\}\times C$ has nowhere dense intersection with every $\overline{Z}_j.$ Thus, $C\times C$ has nowhere dense intersection with every $\overline{Z}_j.$ Therefore, the restriction of $X$ over $C$ is non-isotrivial. Thus, the result is proved by Noetherian induction on $T.$ QED

Example. There are examples where you need to remove a dense open $U$ as in the proof. For instance, let $S$ be the Hilbert scheme of length $4$, zero-dimensional closed subschemes of $\mathbb{P}^2$ that are reduced. Let $X\to S$ be the blowing up of $S\times \mathbb{P}^2$ along the universal closed subscheme of $S\times \mathbb{P}^2.$ This is isotrivial over a dense Zariski open subset of $S$, but it is non-isotrivial over the locus parameterizing linearly degenerate closed subschemes.

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    $\begingroup$ For the existence of the Hilbert scheme, you need $X$ to be projective over $S$. Otherwise, the Hilbert functor is only an algebraic space, but I believe the argument works just as well. $\endgroup$ – Laurent Moret-Bailly Nov 2 '17 at 15:49
  • $\begingroup$ @LaurentMoret-Bailly. Thank you. You are completely correct. I will revise the answer. $\endgroup$ – Jason Starr Nov 2 '17 at 18:31

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