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Background: Properness is a much more robust notion than projectiveness. For example, properness descends along arbitrary fpqc covers (see, for example, Vistoli's Notes on Grothendieck topologies, fibered categories and descent theory, Proposition 2.36). This is far from true for projectiveness. In fact, projectiveness doesn't even descend along Zariski covers! The standard example of a proper non-projective morphism is locally projective over the base. (pictured in an appendix of Hartshorne; page 443, unfortunately not on Google books)

So how delicate is projectivity? Suppose X is a scheme over a field k, and suppose K is a field extension of k such that XK is projective over K. Does it follow that X is projective over k?

The obvious thing to do (I think) is to pick a very ample line bundle L on XK and try to descend it to X. If K is a finite extension of k, then the norm of L will be an ample line bundle on X (if I haven't misunderstood EGA II, section 6.5 and Corollary 6.6.2). But could it be that there is an infinite extension K of k such that XK is projective over K, but X is not projective over k?

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For finite extensions, this is stated explicitly as Corollary 6.6.5 in EGA II and it is also stated there that the result is true for arbitrary extensions. One may reduce the general case to the finite case as follows:

First, we may asume that K is finitely generated since any projective scheme is defined by finitely many equations. This allows us to find a finitely generated integral domain A over k with quotient K and a projective scheme Y over Spec(A) with generic fibre equal isomorphic to X_K. Let Y' = X x Spec(A) and p the projection from Y' to Spec(A). By construction the generic fibre of the two schemes above over Spec(A) are isomorphic. By properness it easily follows that there is an open set U of Spec(A) such that the two schemes become isomorphic over U. U has rational points over a finite extensions of k so we reduce to the finite extension case.

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  • $\begingroup$ Why does properness imply that there's a dense open over which the two are isomorphic? $\endgroup$
    – Anton Geraschenko
    Commented Oct 21, 2009 at 6:01
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    $\begingroup$ Since the generic fibres are isomorphic, there are open sets V and V' of Y and Y' to which any isomorphism extends. The complements of these, say Z and Z', are closed sets whose images in Spec(A) are closed (by properness). Since the generic point is not contained in the image, it follows that the complement in Spec(A) is non-empty and open. (Properness is not really essential; one can use constructibility of images instead.) $\endgroup$
    – naf
    Commented Oct 21, 2009 at 6:40
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The entirety of sections 8--12 (apart from 10) of EGA IV is devoted to fleshing out in remarkably exhaustive and elegant generality the entire yoga of "spreading out and specialization" of which this question is but a special case. Highly recommended reading; it is used (implicitly, if not explicitly) all the time when people prove theorems in algebraic geometry by specialization. This includes proving results over $\mathbf{C}$ by "reduction to the case of positive characteristic or finite fields" (e.g., Mori, Deligne-Illusie) as well as construction of moduli spaces of stable curves by digging out subschemes of Hilbert schemes, etc.

Here is a nifty little exercise to test one's understanding of the EGA formalism: if $X$ and $Y$ are schemes of finite type over a field $k$ and if there is an extension field $K/k$ such that there is a $K$-morphism $f:X _K \rightarrow Y _K$ with any "reasonable" property $\mathbf{P}$ then there is such a morphism with $K/k$ a finite extension; here, "reasonable" can be lots of things: isomorphism, surjective, open immersion, closed immersion, finite flat of degree 42, a semistable curve fibration, smooth, proper and flat with geometric fibers having 12 irreducible components which intersect according to such-and-such configuration and dimensions, and so on. The point is that the initial $f$ is certainly not descended to a finite subextension of the initial $K/k$, and if you made the construction over such an extension and extended scalars back up to the original $K$ then it has absolutely nothing to do with the original $f$.

On the topic of specialization for morphisms, I can't resist mentioning a useful fact which is not a formal consequence of that general stuff: if $A$ and $B$ are abelian varieties over a field $k$ then there exists a finite (even separable) extension $k'/k$ such that (loosely speaking) "all homomorphisms from $A$ to $B$" are defined over $k'$. This means that if $K/k'$ is any extension field whatsoever, then every $K$-homomorphism $A_K \rightarrow B_K$ is defined over $k'$. (Quick proof: the locally finite type Hom-scheme has finitely generated group of geometric points, and is unramified by functorial criterion, so it is \'etale since we are over a field.) There is nothing like this for general (even proper smooth) varieties; just think about automorphisms of projective space.

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    $\begingroup$ Another proof of the abelian variety fact: Each homomorphism is (loosely speaking, again) defined over a separable closure k^s since it maps a Zariski dense subset of A(k^s) into B(k^s), namely the prime-to-the-characteristic torsion points. Since the group Hom(A,B) is finitely generated, this is enough. $\endgroup$
    – Bjorn Poonen
    Commented Feb 16, 2010 at 5:50
  • $\begingroup$ In fact, it suffices to take k' = k(A[3]) (when char(k) is not 3): see Fields of definition for homomorphisms of abelian varieties, A. Silverberg, Journal of Pure and Applied Algebra 77 (1992), 253-262. $\endgroup$
    – Bjorn Poonen
    Commented Feb 16, 2010 at 5:52
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    $\begingroup$ Getting to a finite algebraic extension requires only the elementary 16.9 of my algebraic geometry notes. Silverberg uses a theorem of Chow to get to a separable extension. Here's a modern elementary argument. We have to show that an abelian variety $A$ over $k$ doesn't acquire extra endomorphisms when we pass from $k$ to $k[a]$, $a^p\in k$. By descent for $k\to k[a]$, it suffices to check that $A$ doesn't acquire extra endomorphisms when we pass from $k$ to $k[e]$, $e^p=0$, but this follows from (an almost trivial case of) the rigidity lemma. $\endgroup$
    – JS Milne
    Commented Feb 16, 2010 at 12:57
  • $\begingroup$ This is a very nice discussion. I wanted to insert a warning about characteristic p weirdness (quoted from memory; please correct if necessary): on the other hand, there is an abelian variety A over an algebraically closed field of characteristic p and a positive integer N (= p^a for some a, presumably) such that the space of order N closed subgroup schemes of A varies in moduli: in particular, you get more by enlarging the algebraically closed base field. $\endgroup$
    – Pete L. Clark
    Commented Feb 16, 2010 at 13:59
  • $\begingroup$ I personally like Jim's argument best of all, but I decided against using that descent/rigidity method above since the Hom-scheme argument fit in one sentence. Pete, for your comment see Theorem 3.18 in my expository paper on Chow trace. $\endgroup$
    – BCnrd
    Commented Feb 17, 2010 at 2:53

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