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$\textbf{Question 1}$ (Generically projective over $\bar{k}$) Let $X, Y$ be schemes of finite type over a fixed field $k = \bar{k}$, and we work with $k$-schemes. Suppose $\phi: Y \rightarrow X$ is flat of finite type, with $X$ integral and $Y$ proper, irreducible. If the fibers $Y_z$, of $\phi$ over closed points $z$ in a nonempty open subset of $X$ are projective (over $k$), then does it follow that all fibers over closed points are in fact projective (over $k$).

$\textbf{Question 2}$ (Spreading out) If $\phi: Y \rightarrow \textrm{Spec } \mathbb{Z}$ is flat of finite type with $Y$ irreducible and $X$ integral, and the generic fiber $Y_\mathbb{Q} \rightarrow \mathbb{Q}$ a projective morphism. Are the fibers $Y_{\textbf{F}_p} \rightarrow \textrm{Spec } \mathbb{F}_p$ also projective morphisms? Do we have a version of "spreading out", that is, if the morphism is projective at the generic fiber then is is so over a nonempty open set.

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Regarding 1), I guess you want to assume that $\phi$ is proper. Otherwise you could just remove one point from $Y$... –  Piotr Achinger Nov 3 '12 at 2:21
    
Oops, Thanks! –  LMN Nov 3 '12 at 2:24
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If the answer to $1$ were positive then the fact that the generic fibre of $\phi$ is projective and (consequently) has a Néron-Severi group of rank $>0$ would imply that for every fibre of $\phi$, the Néron-Severi group is also of rank $>0$. This is not very likely to be true (I think). –  Damian Rössler Nov 3 '12 at 20:59
    
I suspect that a counterexample to (1) can be obtained as follows: Hironaka's example of a complete smooth nonprojective variety is obtained by taking an arbitrary nonsingular projective three-fold $X$, and performing certain "local" blowups that are glued together in a funny way (see Hartshorne A.3.4.1) to obtain the desired nonprojective variety $\tilde{X}$. I believe that the construction for "deformation to the normal cone" can be modified to produce a flat, proper family over $\mathbb P^1$ such that the general fiber is $X$, but one special fiber is reducible and ... –  Charles Staats Nov 3 '12 at 21:31
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... one irreducible component of this special fiber is isomorphic to $\tilde{X}$. Consequently, the special fiber cannot be projective, even though the general fiber is the projective variety $X$. –  Charles Staats Nov 3 '12 at 21:33

1 Answer 1

up vote 4 down vote accepted

For question 2: Let $Y\to S$ be a proper morphism with $S$ integral and noetherian (for simplicity) and $Y$ irreducible with non-empty projective generic fiber $Y_K$ (where $K$ is the field of rational functions on $S$). Then there exists a dense open subset $U$ of $S$ such that $Y_U\to U$ is projective:

Let $Y_K\to \mathbb P^N_K$ be a closed immersion. Let $Z$ be the scheme-theoretical closure of the image $Y_K$ in $\mathbb P^N_K$. Then we have a birational map $f$ from $Y$ to $Z$. The domain of definition $\Omega$ of $f$ is open and contains the generic fiber $Y_K$. So $\Omega$ contains $Y_V$ for some dense open subset $V$ of $S$ and we have a birational morphism $Y_V\to Z_V$.

Applying the same reasonning to this morphism or to the inverse birational map, we see that there exists a dense open subset $U$ of $V$ such that $Y_U\to Z_U$ is an isomorphism. So $Y_U\to U$ is projective.

Now for all $s\in U$, $Y_s$ is projective over the residue field $k(s)$ of $S$ at $s$.

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You can also invoke EGA IV, (8.10.5)(xiii). –  Laurent Moret-Bailly Nov 5 '12 at 7:25

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